LeetCode 3Sum 双指针
作者:互联网
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.
Solution
我们需要输出方案,不在乎下标,但是不能有重复的。不妨先排序,然后利用遍历第一个选择 \(nums[i]\),剩下的 \(j,k\) 利用双指针(从 \(l=i+1,r=n-1\) 开始),如果此时的 \(sum>0\),意味着右指针左移 \(r-=1\);否则左指针右移 \(l+=1\)
点击查看代码
class Solution {
private:
vector<vector<int>> ans;
public:
vector<vector<int>> threeSum(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(),nums.end());
for(int i=0;i<n-2;i++){
int l = i+1, r = n-1;
if(i!=0 && nums[i]==nums[i-1])continue;
while(l<r){
vector<int> tmp;
int ck = nums[i]+nums[l]+nums[r];
if(ck==0){
tmp.push_back(nums[i]);
tmp.push_back(nums[l]);
tmp.push_back(nums[r]);
ans.push_back(tmp);
// remove duplicated
while(r-1>=l && nums[r-1]==nums[r])r--;
while(l+1<r && nums[l+1]==nums[l])l++;
r--;l++;
}
else if(ck>0) r--;
else l++;
}
}
return ans;
}
};
标签:tmp,nums,3Sum,back,int,push,LeetCode,指针 来源: https://www.cnblogs.com/xinyu04/p/16464623.html