leetcode 15. 3Sum 双指针法
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
先排序,然后依次枚举第一个元素,第二个和第三个元素的指针分别位于后面数组中的开头和末尾,根据三个的和来确定指针的移动:
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(),nums.end());
int sz=nums.size();
vector<vector<int>> res;
for(int i=0;i<sz-2;i++){
int target=-nums[i];
int l=i+1,r=sz-1;
while(l<r){
int v=nums[l]+nums[r];
if(v==target){
res.push_back(vector<int>{nums[i],nums[l],nums[r]});
l++,r--;
while(l<r&&nums[r]==nums[r+1])r--;
while(l<r&&nums[l]==nums[l-1])l++;
}
if(v>target){
r--;
while(l<r&&nums[r]==nums[r+1])r--;
}
if(v<target){
l++;
while(l<r&&nums[l]==nums[l-1])l++;
}
}
while(i<sz-2&&nums[i+1]==nums[i])i++;
}
return res;
}
};