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  • 双指针法数组三数之和

    15. 三数之和 - 力扣(LeetCode) (leetcode-cn.com)

     1 class Solution {
     2 public:
     3     vector<vector<int>> threeSum(vector<int>& nums) {
     4         vector<vector<int>> result;
     5         sort(nums.begin(), nums.end());
     6     
     7         // 找出a + b + c = 0    a = nums[i], b = nums[left], c = nums[right]
     8         for (int i = 0; i < nums.size(); i++) {   //去重原则,先判断了,利用了条件,再下一位碰到同样元素时去重
     9             // 对a去重
    10             if(i > 0 && nums[i] == nums[i - 1]) {
    11                 continue;
    12             }
    13             int left = i + 1;
    14             int right = nums.size() - 1;
    15             while (right > left) {
    16            
    17                 if (nums[i] + nums[left] + nums[right] > 0) {
    18                     right--;
    19                 } else if (nums[i] + nums[left] + nums[right] < 0) {
    20                     left++;
    21                 } else if (nums[i] + nums[left] + nums[right] == 0){
    22                     result.push_back(vector<int>{nums[i], nums[left], nums[right]});
    23                     right--;
    24                     left++;
    25                     while (right > left && nums[right] == nums[right + 1]) right--;//对b去重
    26                     while (right > left && nums[left] == nums[left - 1]) left++;//对c去重
    27                 }
    28             }
    29 
    30         }
    31         return result;
    32     }
    33 };
    View Code

    18. 四数之和 - 力扣(LeetCode) (leetcode-cn.com)

     1 class Solution {
     2 public:
     3     vector<vector<int>> fourSum(vector<int>& nums,int target) {
     4         vector<vector<int>> result;
     5         sort(nums.begin(), nums.end());
     6         for(int k=0;k<nums.size();k++){
     7             if(k > 0 && nums[k] == nums[k - 1]) {//[2,2,2,2,2]
     8                 continue;
     9             }
    10             for (int i = k+1; i < nums.size(); i++) { 
    11                 if(i>k+1 &&nums[i] == nums[i - 1]) {//i>k+1
    12                     continue;
    13                 }
    14                 int left = i + 1;
    15                 int right = nums.size() - 1;
    16                 while (right > left) {
    17                     // nums[k] + nums[i] + nums[left] + nums[right] > target 会溢出
    18                     if (nums[k] + nums[i] > target - (nums[left] + nums[right])) {
    19                         right--;
    20         
    21                     } else if (nums[k] + nums[i]  < target - (nums[left] + nums[right])) {
    22                         left++;
    23                     } else if (nums[k]+nums[i] + nums[left] + nums[right] == target){
    24                         result.push_back(vector<int>{nums[k],nums[i], nums[left], nums[right]});
    25                         right--;
    26                         left++;
    27                         while (right > left && nums[right] == nums[right + 1]) right--;//对b去重
    28                         while (right > left && nums[left] == nums[left - 1]) left++;//对c去重
    29                     }
    30                 }
    31             }
    32         }
    33         return result;
    34     }
    35 };
    View Code
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  • 原文地址:https://www.cnblogs.com/hcl6/p/15801505.html
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