Remove Duplicate
这道题目和1081. Smallest Subsequence of Distinct Characters是同一道题目，解题方法如下：
class Solution {
public:
string removeDuplicateLetters(string s) {
vector<int> dict(256, 0);
vector<bool> visited(256, false);
for(auto ch : s) dict[ch]++;
string result = "0";
/** the key idea is......
39. Combination Sum
这道题目应当是一类题目，这类题目有个统一的解法—回溯法(backtracking)
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> combination;
sort(candidates.begin(),candidates.end());
co......
315. Count of Smaller Numbers After Self
常规暴力破解解法：
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
int n = nums.size();
vector<int> res;
for(int i = 0;i < n;i++)
{
int cnt = 0;
for(int j = i +1;j < n;j++)
{
if(nums[j] < nums[i])
cnt++;
}
res.......
1572.Matrix Diagonal Sum
常规解法如下，目前没发现有更好的方法，时间复杂度优于O(N).
class Solution {
public:
int diagonalSum(vector<vector<int>>& mat) {
int n = mat.size(),sum = 0;
for(int i = 0;i < n;i++)
{
sum += mat[i][i] + mat[i][n -1 -i];
}
//sub middle element,when n is odd,it add twice.
if(n......
48. Rotate Image
看了大神的这类翻转问题通用解法，
解这类问题就变得非常简单了。
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
reverse(matrix.begin(),matrix.end());
for(int i = 0;i < matrix.size();i++)
{
for(int j = i + 1;j < matrix[i].size();j++)
{
swap(matrix[i][j],matrix[j][i]);
......
1534 Count Good Triplets
无脑上暴力破解解法：
class Solution {
public:
int countGoodTriplets(vector<int>& arr, int a, int b, int c) {
int res = 0,n = arr.size();
for(int i = 0;i < n;i++)
{
for(int j = i + 1;j < n;j++)
{
for(int k = j + 1;k < n;k++)
{
if(abs(arr[i] - arr[j]) <= a &......
Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.
A subarray is a contiguous subsequence of the array.
Return the sum of all odd-length subarrays of arr.
Example 1:
Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and t......
Balanced strings are those who have equal quantity of 'L' and 'R' characters.
Given a balanced string s split it in the maximum amount of balanced strings.
Return the maximum amount of splitted balanced strings.
Example 1:
Input: s = "RLRRLLRLRL"
Output: 4
Explanation: s can be split into "RL", "RR......
Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].
Return the answer in an array.
Example 1:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
E......
常规解法，空间复杂度时间复杂度O(n)
class Solution {
public:
string restoreString(string s, vector<int>& indices) {
string res(s);
int n = s.size();
for(int i = 0;i < n;i++)
{
res[indices[i]] = s[i];
}
return res;
}
};
运行结果：
Runtime: 8 ms, faster than 98.74% of C++ online submissions for Shuf......