LeetCode Online Judge 1. Two Sum

刷个题，击败0.17%...

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

 code:

 1 public class Solution {
 2     public int[] TwoSum(int[] nums, int target) {
 3             int[] result=null;
 4             int i;
 5             for (i = 0; i < nums.Length; i++)
 6             {
 7                 int j;
 8                 for (j = 0; j < nums.Length; j++)
 9                 {
10                     if(i != j)
11                     {
12                         if (nums[i] + nums[j] == target)
13                         {
14                             result = new int[] {j, i};
15                         }
16                     }
17                 }
18             }
19            return result;
20     }
21 }

 

修改一下：

3.63%了...

 1     public int[] TwoSum(int[] nums, int target) {
 2             int[] result=null;
 3             int i;
 4             bool finishLoop = false;
 5             for (i = 0; i < nums.Length; i++)
 6             {
 7                 int j;
 8                 for (j = 0; j < nums.Length; j++)
 9                 {
10                     if(i != j)
11                     {
12                         if (nums[i] + nums[j] == target)
13                         {
14                             result = new int[] {i, j};
15                             finishLoop = true;
16                             break;
17                         }
18                     }
19                 }
20                 if(finishLoop == true)
21                     break;
22             }
23            return result;
24     }

 再改一下：

7.26%

 1             int[] result = null;
 2             int i;
 3             bool finishLoop = false;
 4             for (i = 0; i < nums.Length; i++)
 5             {
 6                 int j;
 7                 for (j = 0; j < nums.Length; j++)
 8                 {
 9                     if (i != j)
10                     {
11                         if (nums[i] + nums[j] == target)
12                         {
13                             result = new[] { i, j };
14                             finishLoop = true;
15                             break;
16                         }
17                     }
18                 }
19                 if (finishLoop)
20                     break;
21             }
22             return result;

试试两个continue：

 1     public int[] TwoSum(int[] nums, int target) {
 2             int[] result = null;
 3             int i;
 4             bool finishLoop = false;
 5             for (i = 0; i < nums.Length; i++)
 6             {
 7                 int j;
 8                 for (j = 0; j < nums.Length; j++)
 9                 {
10                     if (i == j) continue;
11                     if (nums[i] + nums[j] != target) continue;
12                         result = new[] { i, j };
13                         finishLoop = true;
14 
15                 }
16                 if (finishLoop)
17                     break;
18             }
19             return result;
20         }

试试一个continue：

 1     public int[] TwoSum(int[] nums, int target) {
 2             int[] result = null;
 3             int i;
 4             bool finishLoop = false;
 5             for (i = 0; i < nums.Length; i++)
 6             {
 7                 int j;
 8                 for (j = 0; j < nums.Length; j++)
 9                 {
10                     if (i == j) continue;
11                     if (nums[i] + nums[j] == target)
12                     {
13                         result = new[] { i, j };
14                         finishLoop = true;
15                         break;                        
16                     }
17 
18                 }
19                 if (finishLoop)
20                     break;
21             }
22             return result;
23         }

 

posted @ 2017-01-05 20:35  欢崽  阅读(174)  评论(0)  编辑  收藏  举报