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LintCode: 3 Sum

更新时间:2022-09-09 22:57:03

C++

把3个数求和,转变为2个数求和

1. 把数组排序

2. 注意过滤重复值

3. 从前到后遍历,游标i

4. 从后边数中找start + end = -arr[i]的2 sum

5. start + end < -arr[i], start++

6. start + end > -arr[i], end--

7. start + end = -arr[i], insert <i, start, end> into result vecotr

LintCode: 3 Sum
 1 class Solution {
 2 public:    
 3     /**
 4      * @param numbers : Give an array numbers of n integer
 5      * @return : Find all unique triplets in the array which gives the sum of zero.
 6      */
 7     vector<vector<int> > threeSum(vector<int> &nums) {
 8         // write your code here
 9         vector<vector<int> > result;
10         
11         sort(nums.begin(), nums.end());
12         for (int i = 0; i < nums.size(); i++) {
13             if (i > 0 && nums[i] == nums[i - 1]) {
14                 continue;
15             }
16             // two sum;
17             int start = i + 1, end = nums.size() - 1;
18             int target = -nums[i];
19             while (start < end) {
20                 if (start > i + 1 && nums[start - 1] == nums[start]) {
21                     start++;
22                     continue;
23                 }
24                 if (nums[start] + nums[end] < target) {
25                     start++;
26                 } else if (nums[start] + nums[end] > target) {
27                     end--;
28                 } else {
29                     vector<int> triple;
30                     triple.push_back(nums[i]);
31                     triple.push_back(nums[start]);
32                     triple.push_back(nums[end]);
33                     result.push_back(triple);
34                     start++;
35                 }
36             }
37         }
38         
39         return result;
40     }
41 };
LintCode: 3 Sum

 


本文转自ZH奶酪博客园博客,原文链接:http://www.cnblogs.com/CheeseZH/p/5009646.html,如需转载请自行联系原作者