19.1.26 [LeetCode15]3Sum

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]
]

题意

在给定数组中找三个数和为0

题解

用了set储存答案防止重复，很慢

 1 class Solution {
 2 public:
 3     vector<vector<int>> threeSum(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         set<vector<int>> ans;
 6         for (int i = 0; i < nums.size(); i++) {
 7             if (nums[i] > 0)break;
 8             int j = i + 1, k = nums.size() - 1;
 9             while (j < k) {
10                 if (nums[j] + nums[k] == 0 - nums[i]) {
11                     vector<int>tmp;
12                     tmp.push_back(nums[i]);
13                     tmp.push_back(nums[j]);
14                     tmp.push_back(nums[k]);
15                     ans.insert(tmp);
16                     j++,k--;
17                 }
18                 else if (nums[j] + nums[k] < 0 - nums[i])j++;
19                 else k--;
20             }
21         }
22         vector<vector<int>> _ans;
23         while (!ans.empty()) {
24             auto tmp = *(ans.begin());
25             ans.erase(ans.begin());
26             _ans.push_back(tmp);
27         }
28         return _ans;
29     }
30 };

View Code

于是换了一下……

 1 class Solution {
 2 public:
 3     vector<vector<int>> threeSum(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         vector<vector<int>> ans;
 6         for (int i = 0; i < nums.size(); i++) {
 7             if (nums[i] > 0)break;
 8             if (i > 0 && nums[i] == nums[i - 1])continue;
 9             int j = i + 1, k = nums.size() - 1;
10             while (j < k) {
11                 if (nums[j] + nums[k] == 0 - nums[i]) {
12                     vector<int>tmp;
13                     tmp.push_back(nums[i]);
14                     tmp.push_back(nums[j]);
15                     tmp.push_back(nums[k]);
16                     while (j < k&&nums[j] == nums[j + 1])j++;
17                     while (j < k&&nums[k] == nums[k - 1])k--;
18                     ans.push_back(tmp);
19                     j++, k--;
20                 }
21                 else if (nums[j] + nums[k] < 0 - nums[i])j++;
22                 else k--;
23             }
24         }
25         return ans;
26     }
27 };

View Code

差得好多哦-。-

注定失败的战争，也要拼尽全力去打赢它；就算输，也要输得足够漂亮。