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1.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].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.
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/** * Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target) {
int i,j;
int *a = (int *)malloc(sizeof(int) * 2);
for(i=0;i<numsSize;i++){
for(j=i+1;j<numsSize;j++){
if(nums[i]+nums[j]==target){
a[0]=i;
a[1]=j;
break;
}
}
}
//printf("%d",a[1]);
return a;
} |
LeetCode第一题！！！！没想到两层循环就解决了，想想还有点激动。看了网上才知道这样
时间复杂度O（N*2）。
好像快点的话还可以hash表？
有机会再说吧[%>_<%]
本文转自 努力的C 51CTO博客，原文链接:http://blog.51cto.com/fulin0532/1864612