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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. Subscribe to see which companies asked this questio
/**
* 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第一題！！！！沒想到兩層循環就解決了，想想還有點激動。看了網上才知道這樣
時間復雜度O（N*2）。
好像快點的話還可以hash表？
有機會再說吧[%>_<%]
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