t********e
发帖数: 1169 | 1 来源
http://www.foreignlaborcert.doleta.gov/pdf/quarter_2_2012/LCAFY
表格上工资有两项，一个是Employer's proposed wage rate,另一个是max wage rate.
如果max那项值不为零，那就不在统计范围内。
Amazon
SDE1: avg = 96k, num = 42
SDE2: avg = 108k num = 53
SDE3: avg = 133k num = 3
EBay
Soft. Eng. 2 avg = 95k num = 39
Soft. Eng. 3 avg = 111k num = 49
MTS1 Soft Eng. avg = 130k num = 27
MTS2 Soft Eng. avg = 145k num = 2
Facebook
Soft. Eng. avg = 114k num = 115
Google:
Research Eng. avg = 151k num = 10
Research Sci... 阅读全帖 |
|
r******g
发帖数: 13 | 2 O(n^3),
Run Status: Accepted!
Program Runtime: 8 milli secs
Progress: 15/15 test cases passed.
Run Status: Accepted!
Program Runtime: 336 milli secs
my hashtable solution O(n^2) exceeds memory limit, if anyone has code(O^2)
passing OJ, please let me know.
class Solution {
public:
vector > fourSum(vector &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector > results;
int n = num.siz... 阅读全帖 |
|
a**********0
发帖数: 422 | 3 如果有duplicates 用下面这个算法
import java.util.*;
public class PermutationsII {
public ArrayList> permuteUnique(int[] num) {
Arrays.sort(num);
ArrayList> myResult = new ArrayList
Integer>>();
ArrayList temp1 = new ArrayList();
for(int i = 0; i<= num.length-1; i++)
temp1.add(num[i]);
myResult.add(temp1);
while(nextPermutation(num)){
... 阅读全帖 |
|
a**********0
发帖数: 422 | 4 java代码
import java.util.*;
public class PermutationsII {
public ArrayList> permuteUnique(int[] num) {
Arrays.sort(num);
ArrayList> myResult = new ArrayList
Integer>>();
ArrayList temp1 = new ArrayList();
for(int i = 0; i<= num.length-1; i++)
temp1.add(num[i]);
myResult.add(temp1);
while(nextPermutation(num)){
ArrayList<... 阅读全帖 |
|
b******g
发帖数: 3616 | 5 今天正好复习到这题。顺手写了个k-sum的代码，过了LC的3sum和4sum的test。有兴趣帮
着找找有没有没发现的bug。我觉得理解了3sum的算法，并写过4sum以后，这个kSum其
实也就一个小变种，一个葫芦里的药了。由于代码不短，感觉面试的时候还是分成子程
序写比较好，即使时间来不及写完，至少思路能和面试官说清楚。
思路和排板好看点的代码写在自己的复习总结博客里了：
http://bangbingsyb.blogspot.com/2014/11/leetcode-4sum.html
代码：
class Solution {
public:
vector > fourSum(vector &num, int target) {
vector> allSol;
vector sol;
sort(num.begin(),num.end());
kSum(num, 0, num.size()-1, target, 4, sol, allSol);... 阅读全帖 |
|
h***k
发帖数: 161 | 6 把二楼的c++版翻译了一下，能过但应该还能优化
public class Solution {
/**
*@param A: A positive integer which has N digits, A is a string.
*@param k: Remove k digits.
*@return: A string
*/
public String DeleteDigits(String A, int k) {
// write your code here
int len = A.length();
if (len == k) {
return "";
}
int idx = 0;
int[] num = new int[len];
for (int i = 0; i < len; i++) {
num[i] = A.charAt(i) - '0';
... 阅读全帖 |
|
h***k
发帖数: 161 | 7 把二楼的c++版翻译了一下，能过但应该还能优化
public class Solution {
/**
*@param A: A positive integer which has N digits, A is a string.
*@param k: Remove k digits.
*@return: A string
*/
public String DeleteDigits(String A, int k) {
// write your code here
int len = A.length();
if (len == k) {
return "";
}
int idx = 0;
int[] num = new int[len];
for (int i = 0; i < len; i++) {
num[i] = A.charAt(i) - '0';
... 阅读全帖 |
|
c*******0
发帖数: 162 | 8 这个代码O(n^3), 但不超时。
vector > fourSum(vector &num, int target) {
vector > rst;
if(num.size() <= 3) return rst;
std::sort(num.begin(), num.end());
int len = num.size();
for(int i = 0; i < num.size() - 3; i++){
if(i > 0 && num[i] == num[i-1]) continue;
for(int j = i + 1; j < num.size() - 2; j++){
if(j > i + 1 && num[j] == num[j-1])continue;
int l = j + 1, r = len - 1;
... 阅读全帖 |
|
H******7
发帖数: 1728 | 9 public List> fourSum(int[] num, int target) {
ArrayList> ans = new ArrayList<>();
if(num.length<4)return ans;
Arrays.sort(num);
for(int i=0; i
if(i>0&&num[i]==num[i-1])continue;
for(int j=i+1; j
if(j>i+1&&num[j]==num[j-1])continue;
int low=j+1, high=num.length-1;
while(low
int sum=num[i]+num[j]+n... 阅读全帖 |
|
m******3
发帖数: 346 | 10 贴个leetcode讨论区的解答，思想就是一边生成一边计算
void addOperators(vector& result, string nums, string t, long long
last, long long curVal, int target) {
if (nums.length() == 0) {
if (curVal == target)
result.push_back(t);
return;
}
for (int i = 1; i<=nums.length(); i++) {
string num = nums.substr(0, i);
if(num.length() > 1 && num[0] == '0')
return;
string nextNum = nums.substr(i);
if (t.length() > 0) {
addOpera... 阅读全帖 |
|
h*****i
发帖数: 1017 | 11 #include
#include
using namespace std;
class binaryTree {
public:
int value;
int indx;
binaryTree *left;
binaryTree *right;
binaryTree(){
left = NULL;
right = NULL;
}
};
void insert(binaryTree *bTree, int num, int key){
if(bTree == NULL){
bTree = new binaryTree;
bTree->value = num;
bTree->indx = key;
printf("%d %dn",bTree->value,bTree->indx);
return;
... 阅读全帖 |
|
T*******e
发帖数: 4928 | 12 bool noSwap(const vector &num, int k, int i) {
for(int j=k; j
if(num[j]==num[i]) return true; //see if any element from num[k] to
num[i-1] is equal to num[i]. If equal, that means someone equal to num[i]
has already taken this seat before. That possibility has been covered.
}
return false;
}
void permutationUniqueHelper(vector num, int n, int k, vector
int> > &res) {
if(k==n) {
res.push_back(num);
} else {
for(int i=k; i<=n;... 阅读全帖 |
|
G***n
发帖数: 877 | 13 Find three integers in S such that the sum is closest to a given number.
下面的code没问题。我想优化一下，skip那些重复的数，但加上下面这2个while语句然
后返回的结果就错了，搞不懂哪里的问题。
//while (j
// while (j
class Solution {
public:
int threeSumClosest(vector &num, int target) {
int sum = INT_MIN;
int minDist = INT_MAX;
if (num.size()<3) return sum;
sort(num.begin(), num.end());
for (int i = 0; i阅读全帖 |
|
t**r
发帖数: 3428 | 14 class Solution {
public:
vector > fourSum(vector &num, int target) {
int n = num.size();
vector > ret;
sort(num.begin(), num.end());
for (int i = 0; i < n; ++i) {
if (i != 0 && num[i] == num[i - 1]) continue;
for (int j = n - 1; j > i; --j) {
if (j != n - 1 && num[j] == num[j + 1]) continue;
int L = i + 1, R = j - 1;
while (L < R) {
int sum ... 阅读全帖 |
|
发帖数: 1 | 15 根据楼上的思想写了两个版本,不知道对不对
单调栈:
List findMinUnsortedSubArray(int[] nums) {
List res = new ArrayList<>(2);
if (nums == null || nums.length == 0) return res;
int n = nums.length;
Stack s = new Stack<>();
int max = Integer.MIN_VALUE;
s.push(-1);
for (int i=0; i
while (!s.isEmpty() && nums[i] < max && nums[i] < nums[s.peek()]
) {
s.pop();
}
if (nums[i]... 阅读全帖 |
|
l*******0
发帖数: 63 | 16 A solution that could deal with dups. O(N^3).
public ArrayList> fourSum(int[] num, int target) {
int len=num.length;
Arrays.sort(num);
ArrayList> results=new ArrayList
Integer>>();
for(int i=0;i
if(i-1>=0&&num[i]==num[i-1]) continue; //skip dup in outmost
loop
for(int j=i+1;j
if(j-1>=i+1&&num[j]==num[j-1]) continue; //skip dup in 2nd
outmost loop
... 阅读全帖 |
|
S*******C
发帖数: 822 | 17 http://oj.leetcode.com/problems/subsets/
第一种格式：res作为局部变量避免线程安全问题，但比较啰嗦
public ArrayList> subsets(int[] num){
if(num==null) return null;
Arrays.sort(num);
ArrayList> res=new ArrayList>(
);
res.add(new ArrayList());// [[]]
dfs(res, num,0,new ArrayList());
return res;
}
private void dfs(ArrayList> res, int[] num, int pos,
ArrayList temp){... 阅读全帖 |
|
s********x
发帖数: 81 | 18 大家好，最近做了两题leetcode, 但是都遇到了run time error. 但是我遇到run time
error 的例子在别的c++编译器上都顺利通过了，请大家帮我看一下是什么问题. 谢谢
大家!
问题一：next permutation:
class Solution {
public:
int findSecondMax(vector &num, int cur){
int max1=-1, max2=-2;
int index1=-1, index2=-2;
for(int i=cur; i
if(max1
max2=max1;
index2=index1;
max1=num[i];
index1=i;
}
}
if(max2!=-2 || max2!=-1) return index2;
else return index1;
}
... 阅读全帖 |
|
f**********t
发帖数: 1001 | 19 Well, my code is complex and sigfault. Let me go back if I have time.
This is too hard for a 45 minutes interview.
void skipSpace(const char **s) {
while (isspace(**s)) {
++(*s);
}
}
int getNum(const char **s) {
bool neg = false;
if (**s == '-') {
neg = true;
++(*s);
}
int cur = 0;
while (isdigit(**s)) {
cur = cur * 10 + (**s - '0');
++(*s);
}
--(*s);
return neg ? -cur : cur;
}
int opNum(int x, int y, char op) {
switch (op) {
case '+':
return x + y... 阅读全帖 |
|
b**w
发帖数: 78 | 20 Python的，写的比较乱
class Solution(object):
def addOperators(self, num, target):
"""
:type num: str
:type target: int
:rtype: List[str]
"""
result = self.helper(num, 0, target)
return result
def helper(self, num, pos, target):
result = []
if pos>=len(num):
if target==0:
result.append("")
return result
if pos==len(num)-1:
if int(num[pos])==target:
... 阅读全帖 |
|
E*******0
发帖数: 465 | 21 // NextPermutation.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include
#include
using namespace std;
#include
bool myfunction (int i,int j) { return (i>j); }
bool nextPermutation(vector &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
//If vector size is 1, permutation is its self.
if (1==num.size())
return true;
// rit is the iterator indicates current ite... 阅读全帖 |
|
g**G
发帖数: 767 | 22 Did you consider to skip duplicate element?
I pasted my solution as follows, hope it helps
public ArrayList> threeSum(int[] num) {
ArrayList> threeSum = new ArrayList
Integer>>();
if (num == null || num.length == 0)
return threeSum;
Arrays.sort(num);
int len = num.length;
int prev1 = Integer.MIN_VALUE;
int prev2 = Integer.MIN_VALUE;
int prev3 = Integer.MIN_VALUE;
for (... 阅读全帖 |
|
j*****8
发帖数: 3635 | 23 我的28行，不过看着也挺繁琐的。思路就是碰到递增的就一直加1，递减的话一直减，
碰到往上走时就回头check一下看那段递减的需不需要调整（负数，或者最小的不是1）
public int candy(int[] ratings) {
if(ratings == null || ratings.length == 0) return 0;
int len = ratings.length;
int[] num = new int[len];
for(int i = 0; i < len; i++) {
if(i == 0) num[i] = 1;
else {
if(ratings[i] > ratings[i-1]) num[i] = num[i-1] + 1;
else if(ratings[i] == ratings[i-1]) num[i] = 1;
else {
... 阅读全帖 |
|
z****e
发帖数: 54598 | 24 leetcode 不超过两秒就可以过大oj
我用了800毫秒，还有1200毫秒富余
把你的code贴上来看看
Run Status: Accepted!
Program Runtime: 864 milli secs
public class Solution {
public ArrayList> fourSum(int[] num, int target) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList> list = new ArrayList
>>();
Set set = new HashSet>();
Arrays.sort(num);
for(int i=0;i
... 阅读全帖 |
|
k*******3
发帖数: 2 | 25 class Solution:
# @return a list of lists of length 3, [[val1,val2,val3]]
def threeSum(self, num):
if len(num) < 3:
return []
num.sort()
final_results = []
for i in range(len(num)-2):
if i == 0 or num[i] > num[i-1]:
first = num[i]
sum_of_two = 0 - first
left_index = i + 1
right_index = len(num) - 1
while left_index < right_index:
tmp_... 阅读全帖 |
|
a***e
发帖数: 413 | 26 Given a collection of numbers that might contain duplicates, return all
possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].
不清楚为什么这个答案会引起Output Limit Exceeded
而把if (i>k&&num[i]==num[k])
改成一个
bool noswap(int k, int i, const vector num){
for (int j=k;j
if (num[j]==num[i]){
return true;
}
}
return false;
}
就被accept了。请问如何能提高解决这一类问题的能力呢？感觉工作中很少，几... 阅读全帖 |
|
b**********0
发帖数: 7 | 27 int findMin(vector &num) {
int start = 0;
int end = num.size()-1;
while(start<=end){
int mid = (start + end)/2;
if(start>=end -1) { return (num[start]>num[end])?num[end]:num
[start]; }
if(num[start] < num[mid] && num[mid] < num[end]) return num[
start];
if(num[mid]==num[end]) {end --; }
else if(num[mid]
else if(num[mid] > num[start]) {start = mid+1; }
... 阅读全帖 |
|
w********n
发帖数: 4752 | 28 class Solution:
# @return a list of lists of length 3, [[val1,val2,val3]]
def threeSum(self, num):
size = len(num)
if size < 3:
return []
num.sort()
res = []
for i in range(size):
if i >0 and num[i] == num[i-1]:
continue
target = -num[i]
start = i+1
end = size - 1
while start < end:
if target == num[start] + num[end]:
res.appe... 阅读全帖 |
|
p*****2
发帖数: 21240 | 29 import java.io.*;
import java.util.*;
public class Roman
{
public static void main(String[] args)
{
new Roman().run();
}
PrintWriter out = null;
void run()
{
Scanner in = new Scanner(System.in);
out = new PrintWriter(System.out);
String s = in.next();
out.println(romanToInt(s));
out.close();
}
public String intToRoman(int num)
{
StringBuffer sb = new StringBuffer();
sb.append(Convert(num, 1000, ne... 阅读全帖 |
|
q****m
发帖数: 177 | 30 优化了以后，终于可以 pass large 了. 思路是先对 num 排序. 排序后 num[i] + num
[j] + num[k] = 0中只用考虑 i < j < k的情况
class Solution {
public:
vector > threeSum(vector &num) {
vector > *set = new vector >;
sort(num.begin(), num.end());
vector b;
b.reserve(num.size());
for(int k = 0; k < num.size(); ++k)
{
int c = num[k];
for(int j = 0; j < k; ++j)
{
b[j] = -c - num[j];
}
int i = 0;
... 阅读全帖 |
|
q****m
发帖数: 177 | 31 优化了以后，终于可以 pass large 了. 思路是先对 num 排序. 排序后 num[i] + num
[j] + num[k] = 0中只用考虑 i < j < k的情况
class Solution {
public:
vector > threeSum(vector &num) {
vector > *set = new vector >;
sort(num.begin(), num.end());
vector b;
b.reserve(num.size());
for(int k = 0; k < num.size(); ++k)
{
int c = num[k];
for(int j = 0; j < k; ++j)
{
b[j] = -c - num[j];
}
int i = 0;
... 阅读全帖 |
|
l*******b
发帖数: 2586 | 32 用leetcode上next permutation的办法就可以过吧
整数版本的，换成字符就好啦
vector > permute(vector &num) {
sort(num.begin(), num.end());
vector > r;
r.push_back(num);
while(nextPermute(num))
r.push_back(num);
return r;
}
bool nextPermute(vector &num) {
int i,j,n;
i = j = num.size() - 1;
while(i > 0 && num[i] <= num[i-1])
--i;
n = i - 1;
if(n < 0) return false;
else{... 阅读全帖 |
|
p****e
发帖数: 3548 | 33 搞了个复杂的。。
// Type your C++ code and click the "Run Code" button!
// Your code output will be shown on the left.
// Click on the "Show input" button to enter input data to be read (from
stdin).
#include
using namespace std;
bool OpHigherOrder(char c1, char c2)
{
if(c1 != '(' && c2 == ')')
return true;
else if(c1 == '(')
return false;
else if((c1 == '*' || c1 == '/'))
return true;
else if((c1 == '+' || c1 == '-') && (c2 == '+' || c2 == '-'))
... 阅读全帖 |
|
p****e
发帖数: 3548 | 34 搞了个复杂的。。
// Type your C++ code and click the "Run Code" button!
// Your code output will be shown on the left.
// Click on the "Show input" button to enter input data to be read (from
stdin).
#include
using namespace std;
bool OpHigherOrder(char c1, char c2)
{
if(c1 != '(' && c2 == ')')
return true;
else if(c1 == '(')
return false;
else if((c1 == '*' || c1 == '/'))
return true;
else if((c1 == '+' || c1 == '-') && (c2 == '+' || c2 == '-'))
... 阅读全帖 |
|
g********E
发帖数: 178 | 35 1，第一层循环的时候跳过重复的；
2，第二层循环从两头跳过循环的；
vector< vector > threeSum(vector &num){
int n = num.size();
vector< vector > res;
vector cur;
sort(num.begin(), num.end());
for(int i = 0; i < n-2; i++){
if(i > 0 && num[i] == num[i-1]) continue;
cur.push_back(num[i]);
int start = i + 1, end = n - 1;
while(start < end){
if(start > i+1 && num[start] == num[start-1]) {
start++; continue;
... 阅读全帖 |
|
q****m
发帖数: 177 | 36 能把你的贴一下吗？ 这是我的code.
class Solution {
public:
void nextPermutation(vector &num) {
int i = num.size() - 1;
while(i >= 1 && num[i - 1] >= num[i])
{
--i;
}
if(i == 0)
{
int l = 0, r = num.size() -1 ;
while(l <= r)
{
swap(num[l], num[r]);
++l;
--r;
}
return;
}
int left = i, right = num.size() - 1, targe... 阅读全帖 |
|
o***c
发帖数: 32 | 37 我同学两周前onsite被考到了，不过此题线段树写起来也方便。
#define Maxn 10000
int val[1010];
class Segment_Tree {
private:
struct Node {
int left,right,cover;
};
Node Tree[Maxn];
int Number, c, d;
public:
void build(int Now, int a,int b) {
Tree[Now].left=a;
Tree[Now].right=b;
if(b-a>=1) {
int mid=(a+b)>>1;
Tree[Now].left=a;
build(2*Now+1 , a,mid);
Tree[Now].right=b;
build(2*Now+2,mid+1, b);
Tree[Now]... 阅读全帖 |
|
t**r
发帖数: 3428 | 38 我的代码：
很糟糕，大数据过不去
应该可以用bitmap 优化一下 今天懒得弄了
package topcoder;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
//B[0] = 0
//B[i] = A[B[i-1]], for every i between 1 and N-1, inclusive.
/*
* Permutation Child array
{0, 1, 2} {0, 0, 0}
{0, 2, 1} {0, 0, 0}
{1, 0, 2} {0, 1, 0}
{1, 2, 0} {0, 1, 2}
{2, 0, 1} {0, 2, 1}
{2, 1, 0} {0, 2, 0}
* */
public... 阅读全帖 |
|
l******s
发帖数: 3045 | 39 5 seems easier.
Below is Divide and Conquer for 4.(2)
private bool quarterPart(int[] nums)
{
for (int counter = 0, left = 0; counter < 8; counter++)
{
int i = left, j = left + (nums.Length >> 2);
if (j >= nums.Length) return false;
if (nums[i] == nums[j]) return true;
for (int mid = (i + j) >> 1; i < j; left = i, mid = (i + j) >> 1)
if (nums[i] == nums[mid] || nums[i] != nums[mid] && nums[mid] != nums[
j])
i = mid + 1;
else if (nums[mid] == nums[j])
... 阅读全帖 |
|
l******b
发帖数: 39 | 40
俺来班门弄斧一下, 看这样行不行?
为简化起见, 根据OP的要求, 先建两张表,一张放数据, 一张放参数.
CREATE TABLE t7(
CATEGORY INT,
ID INT,
num int
);
INSERT INTO t7(CATEGORY, ID, num) VALUES(1,1,2) ;
INSERT INTO t7(CATEGORY, ID, num) VALUES(1,2,1) ;
INSERT INTO t7(CATEGORY, ID, num) VALUES(1,3,6) ;
INSERT INTO t7(CATEGORY, ID, num) VALUES(1,4,3) ;
INSERT INTO t7(CATEGORY, ID, num) VALUES(1,5,5) ;
INSERT INTO t7(CATEGORY, ID, num) VALUES(2,1,2) ;
INSERT INTO t7(CATEGORY, ID, num) VALU... 阅读全帖 |
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发帖数: 1 | 41 SELECT bus.n1, bus.c1, stops.name, bus.n2, bus.c2 FROM
(
SELECT b1.num n1, b1.company c1, b1.stop stop, b2.num n2, b2.company c2
FROM
(
SELECT route.num num, route.company company, route.stop stop FROM
(SELECT num, company, MIN(pos) pos, stop FROM (SELECT id, name FROM
stops WHERE name = 'Craiglockhart') s1 INNER JOIN route ON s1.id = route.
stop GROUP BY num, company, stop) r1
INNER JOIN route
ON r1.num = route.num AND r1.company = route.company
WHERE route.stop <> r1.... 阅读全帖 |
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j*****g
发帖数: 463 | 42 更新：
瞎蒙出来根据区号，选择性使用哪一 GV account。
但 CID 似乎还没弄出来。如果有大侠，能否帮忙搞定 CallerID，并检查一下语法. 多谢。
原问题和代码如下
==============================
用 SIPGate + GoogleVoice + SIPSorcery, 一直用 Simple Dial plan.
有哪位高人能发个具有如下功能的 Complex Dial plan 么？
1. 根据区号自动选择使用哪一 GV account.
本人有两个 GV account, 一个搬家前的区号，另一搬家后现在的区号。希望和老朋
友联系时用搬家前的号码。打现在本地时，使用新的号码。
2. 来电显示
现在来电显示在姓名处显示的是 SIPGate 的号码，在电话号码处显示的是对方真实号
码。希望能将 SIPGate 号码替换成对方实际姓名。
我知道网上的 Complex Dial plan 可以 Customize 实现上述功能。
想偷懒，如果谁有现成的直接能用最好。
另一方面也是实在忙得没空。
多谢。
张
Update:
=====... 阅读全帖 |
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k*******9
发帖数: 7 | 43 是的，0100100010000111 -> 1110000100010010。
用的以前见过的这个算法：
Num = (Num & 0x55555555) << 1 | (Num >> 1) & 0x55555555;
Num = (Num & 0x33333333) << 2 | (Num >> 2) & 0x33333333;
Num = (Num & 0x0F0F0F0F) << 4 | (Num >> 4) & 0x0F0F0F0F;
Num = (Num & 0x00FF00FF) << 8 | (Num >> 8) & 0x00FF00FF;
Num = (Num & 0x0000FFFF) << 16 | (Num >> 16) & 0x0000FFFF;
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C*******n
发帖数: 24 | 44 标记个start，然后一步一步开始做
java code，貌似我童鞋也被考这个题了。
个人做了下，蛮有思路，耗时有点长，但是差不多能handle
不过真的算是个难的phone题啊，我这个小弱是这么觉得的。。。。
public void solution(int num){
int i = 1;
while(i * i <= num){
if(i == 1){
System.out.println(num + " * 1");
}else if( num % i == 0){
ArrayList> tmp = helper(num / i, i);
for (ArrayList r : tmp){
for ( Integer no : r){
System.out.print(no + " * ");
}
System.out.println(i);
... 阅读全帖 |
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h*****7
发帖数: 60 | 45 试了 也过不了 如下：
class Solution:
# @return a list of lists of length 3, [[val1,val2,val3]]
def threeSum(self, num):
num.sort()
n = len(num)
res= []
if n<3:
return res
for i in range(n-2):
j = n-1
if i > 0 and num[i] == num[i-1]:
continue
k = i + 1
while k < j:
sum = num[i] + num[j] + num[k]
if sum < 0:
k += 1
el... 阅读全帖 |
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G***n
发帖数: 877 | 46 总是有Output Limit Exceeded的error怎么回事？
class Solution {
public:
vector > threeSum(vector &num) {
vector> list;
if (num.size() <= 2) return list;
sort(num.begin(), num.end());
for (int i = 0; i
{
int j = i+1, k=num.size()-1;
while (j
{
int t = num[i]+num[j]+num[k];
if (t == 0)
{
... 阅读全帖 |
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b*******w
发帖数: 56 | 47 def rotate_by_k(matrix, k):
i = 0
while i < len(matrix)>>1:
# construct 1-d array
nums = []
x, y = i, i
while x < len(matrix) - i - 1:
nums.append(matrix[y][x])
x += 1
while y < len(matrix) - i - 1:
nums.append(matrix[y][x])
y += 1
while x > i:
nums.append(matrix[y][x])
x -= 1
while y > i:
nums.append(matrix[y][x])
y -= 1
... 阅读全帖 |
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c*********i
发帖数: 46 | 48 int findKthLargest(vector& nums, int k) {
return findKth(nums, k, 0, nums.size()-1);
}
int findKth(vector& nums, int k, int start, int end)
{
if(start==end)
return nums[start];
int pos = partition(nums, start, end);
int size = pos-start+1;
if(k==size)
return nums[pos];
else if(k>size)
return findKth(nums, k-size, pos+1, end);
else
return findKth(nums, k, start, po... 阅读全帖 |
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B********t
发帖数: 147 | 49 写了下google第二题
bool IsSumEqualTarget(int target, int i, vector &nums, vector
int>> &sum) {
if (!target && i == nums.size()) return true;
for (int j = i; j < nums.size(); ++j) {
if(IsSumEqualTarget(target-sum[i][j], j+1, nums, sum)) return true;
}
return false;
}
bool IsSumEqualTarget(int target, vector &nums) {
if (nums.empty()) return false;
vector> sum(nums.size(), vector(nums.size(), 0));
sum[0][0] = nums[0];
for (int i ... 阅读全帖 |
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n****5
发帖数: 81 | 50 用 C 写了一下，用的递归来处理商和余数。用的unsigned int所以假定输入小于1百万
X1百万
#include
#include
const char* tens[] = {"Twenty", "Thirty", "Forty", "Fifty", "Sixty", "
Seventy", "Eighty", "Ninety"};
const char* lt20[] = {"Zero", "One", "Two", "Three", "Four", "Five", "Six",
"Seven", "Eight", "Nine", "Ten",
"Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "
Sixteen", "Seventeen", "Eighteen", "Nineteen"};
void num2str(unsigned int num)
{
if (num == 0)
return;
el... 阅读全帖 |
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