algorithm/leetcode-master
1
0
Fork 0
mirror of https://github.com/youngyangyang04/leetcode-master.git synced 2025-07-06 23:28:29 +08:00
Code Issues Packages Projects Releases Wiki Activity

Update

Browse Source
This commit is contained in:
youngyangyang04
2020-07-06 09:13:14 +08:00
parent 87c2efb822
commit abd4aa6c0d
1 changed files with 5 additions and 13 deletions
Download Patch File Download Diff File Expand all files Collapse all files

18
problems/0015.三数之和.md
View File

@ -24,25 +24,27 @@ public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
// 找出a + b + c = 0
// a = nums[i], b = nums[j], c = -(a + b)
for (int i = 0; i < nums.size(); i++) {
// 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
if (nums[i] > 0) {
continue;
}
if (i > 0 && nums[i] == nums[i - 1]) { //三元组第一个元素去重
if (i > 0 && nums[i] == nums[i - 1]) { //三元组元素a去重
continue;
}
unordered_set<int> set;
for (int j = i + 1; j < nums.size(); j++) {
if (j > i + 2
&& nums[j] == nums[j-1]
&& nums[j-1] == nums[j-2]) { // 三元组第三个元素去重
&& nums[j-1] == nums[j-2]) { // 三元组元素b去重
continue;
}
int c = 0 - (nums[i] + nums[j]);
if (set.find(c) != set.end()) {
result.push_back({nums[i], nums[j], c});
set.erase(c);// 三元组第二个元素去重
set.erase(c);// 三元组元素c去重
} else {
set.insert(nums[j]);
}
@ -52,16 +54,6 @@ public:
}
};
```
```
Input
[-2,0,0,2,2]
Output
[[-2,2,0],[-2,2,0]]
Expected
[[-2,0,2]]
```
## 双指针法代码
```
class Solution {