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  • 18. 4Sum(双指针)



    Given an array nums of n integers and an integer target, are there elements abc, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

    Note:

    The solution set must not contain duplicate quadruplets.

    Example:

    Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
    
    A solution set is:
    [
      [-1,  0, 0, 1],
      [-2, -1, 1, 2],
      [-2,  0, 0, 2]
    ]

     1 class Solution {
     2 public:
     3     vector<vector<int>> fourSum(vector<int>& a, int target) {
     4     vector<vector<int>> res;
     5     const int n = a.size();
     6     if (n < 4) return res;
     7     std::sort(a.begin(),a.end());
     8     for (int i = 0; i < n-3; ++i) {
     9         if(i!=0 && a[i]==a[i-1]) {continue;} //去重
    10         for (int j = i+1; j < n-2; ++j) {
    11             if(j!=i+1 && a[j]==a[j-1]){continue;} //去重
    12             int low = j+1;
    13             int high = n-1;
    14             while (low < high) {
    15                 int sum = a[i]+a[j]+a[low]+a[high];
    16                 if (sum > target) {
    17                     high--;
    18                 } else if (sum < target) {
    19                     low++;
    20                 } else {
    21                     vector<int> tep = {a[i],a[j],a[low],a[high]};
    22                     res.emplace_back(tep);
    23                     while(low <high && a[low]==a[low+1]) {low++;}//去重
    24                     while(low <high && a[high]==a[high-1]) {high--;}//去重
    25                     low++;
    26                     high--;
    27                 }
    28             }
    29             
    30         }
    31     }
    32     return res;
    33     }
    34 };


     1 class Solution {
     2     public List<List<Integer>> fourSum(int[] nums, int target) {
     3         List<List<Integer>> res = new LinkedList<>();
     4         if (nums.length<4) return res;
     5         Arrays.sort(nums);
     6         for(int i=0;i<nums.length-3;i++){
     7             if(i>0&&nums[i]==nums[i-1]) continue;
     8             for(int j=i+1;j<nums.length-2;j++){
     9                 if(j>i+1&&nums[j]==nums[j-1]) continue;
    10                 
    11                 int lo = j+1,hi = nums.length-1;
    12                 while(lo<hi){
    13                     int sum = nums[i]+nums[j]+nums[lo]+nums[hi];
    14                     if(sum==target){
    15                         res.add(Arrays.asList(nums[i],nums[j],nums[lo],nums[hi]));
    16                        
    17                         //答案去重
    18                         while(lo<hi&&nums[lo]==nums[lo+1])  lo++;
    19                         while(lo<hi&&nums[hi]==nums[hi-1])  hi--;
    20                         
    21                         lo++;
    22                         hi--;
    23                     }
    24                     
    25                     else if(sum<target) lo++;
    26                     else hi--;
    27                 }
    28                
    29             }
    30         }
    31         return res;
    32     }
    33 }
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  • 原文地址:https://www.cnblogs.com/zle1992/p/9314157.html
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