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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[%>_<%]
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