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:

 public class Solution {
public int[] TwoSum(int[] nums, int target) {
int[] result=null;
int i;
for (i = ; i < nums.Length; i++)
{
int j;
for (j = ; j < nums.Length; j++)
{
if(i != j)
{
if (nums[i] + nums[j] == target)
{
result = new int[] {j, i};
}
}
}
}
return result;
}
}

修改一下:

3.63%了...

     public int[] TwoSum(int[] nums, int target) {
int[] result=null;
int i;
bool finishLoop = false;
for (i = ; i < nums.Length; i++)
{
int j;
for (j = ; j < nums.Length; j++)
{
if(i != j)
{
if (nums[i] + nums[j] == target)
{
result = new int[] {i, j};
finishLoop = true;
break;
}
}
}
if(finishLoop == true)
break;
}
return result;
}

再改一下:

7.26%

             int[] result = null;
int i;
bool finishLoop = false;
for (i = ; i < nums.Length; i++)
{
int j;
for (j = ; j < nums.Length; j++)
{
if (i != j)
{
if (nums[i] + nums[j] == target)
{
result = new[] { i, j };
finishLoop = true;
break;
}
}
}
if (finishLoop)
break;
}
return result;

试试两个continue:

     public int[] TwoSum(int[] nums, int target) {
int[] result = null;
int i;
bool finishLoop = false;
for (i = ; i < nums.Length; i++)
{
int j;
for (j = ; j < nums.Length; j++)
{
if (i == j) continue;
if (nums[i] + nums[j] != target) continue;
result = new[] { i, j };
finishLoop = true; }
if (finishLoop)
break;
}
return result;
}

试试一个continue:

     public int[] TwoSum(int[] nums, int target) {
int[] result = null;
int i;
bool finishLoop = false;
for (i = ; i < nums.Length; i++)
{
int j;
for (j = ; j < nums.Length; j++)
{
if (i == j) continue;
if (nums[i] + nums[j] == target)
{
result = new[] { i, j };
finishLoop = true;
break;
} }
if (finishLoop)
break;
}
return result;
}

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