Description
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, and you may not use the same element twice.Translation
整形数组中每个元素均不相同，且两两相加的结果也不相同。给定某一目标数字，返回数组中相加结果为给定目标数字的两个元素的下标。
Example
Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].
Solution
/** * Note: The returned array must be malloced, assume caller calls free(). */
Method1 (循环遍历)
对数组内的元素两两逐次相加，记录其下标。
int* twoSum(int* nums, int numsSize, int target) { int* arrIndices = (int*)malloc(sizeof(int) * 2); for(i = 0; i < numsSize - 1; i++){ for(j = i + 1; j < numsSize; j++){ if (nums[i] + nums[j] == target){ flag[0] = i; flag[1] = j; break; } else continue; } } return arrIndices;} 时间复杂度： O(n2)
空间复杂度： O(1)