LEETCODE ALGORITHM:15. 3Sum

发表于 更新于 分类于 Waline： 阅读次数：

题目

    • Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

      Notice that the solution set must not contain duplicate triplets.

      Example 1:

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      Input: nums = [-1,0,1,2,-1,-4]
      Output: [[-1,-1,2],[-1,0,1]]

      Example 2:

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      Input: nums = []
      Output: []

      Example 3:

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      Input: nums = [0]
      Output: []

      Constraints:

      • 0 <= nums.length <= 3000
      • -105 <= nums[i] <= 105

题解

i,j,k分别代表排序后数组里三个数的索引号

因为要避免重复，一样的数组不要，所以如果任何一个位置前后两次选择的数一致就要跳过

第三个数的指针倒叙来找

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n=nums.size();
        if(n==0) return {};
        vector<vector<int>> ans;
        sort(nums.begin(),nums.end());
        for(int i=0;i<n;i++){
            if(i>0 && nums[i]==nums[i-1]){
                continue;
            }
            int target= -nums[i];
            int k=n-1;
            for(int j=i+1;j<n;j++){
                if(nums[i]>0) break;
                if(j>i+1 && nums[j]==nums[j-1]){
                    continue;
                }
                while(j<k && nums[j]+nums[k] > target){
                    k--;
                }
                if(j==k) break;
                if(nums[j]+nums[k]==target){
                    ans.push_back(vector<int>{nums[i],nums[j],nums[k]});
                }
            }
        }
        return ans;
    }
};