面试经典150题——Day29

一、题目

15. 3Sum

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:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

3 <= nums.length <= 3000
-105 <= nums[i] <= 105

题目来源： leetcode

二、题解

利用双指针思路解题，关键在于去重，对i，left，right分别进行去重。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        for(int i = 0;i < n;i++){
            if(nums[i] > 0) return res;
            if(i > 0 && nums[i] == nums[i-1]) continue;
            int left = i + 1,right = n - 1;
            while(left < right){
                int tmp = nums[i] + nums[left] + nums[right];
                if(tmp > 0) right--;
                else if(tmp < 0) left++;
                else{
                    vector<int> ve;
                    ve.push_back(nums[i]);
                    ve.push_back(nums[left]);
                    ve.push_back(nums[right]);
                    res.push_back(ve);
                    while(left < right && nums[left] == nums[left+1]) left++;
                    while(left < right && nums[right-1] == nums[right]) right--;
                    left++;
                    right--;
                }
            }
        }
        return res;
    }
};