1. Two Sum

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    int *result = (int*)malloc(2 * sizeof(int));
    
    for(int i = 0; i < numsSize-1; i++){
        for(int j = i+1; j < numsSize; j++){
            if((nums[i] + nums[j]) == target){
                result[0] = i;
                result[1] = j;
            }
        }
    }
    return result;
    
}

注意使用malloc为指针申请空间

 

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        map= {}
        for i,n in enumerate(nums):
            if map.has_key(n):
                return map[n],i
            else:
                map[target - n] = i

map为字典

enumerate是枚举法，i 为索引，n为内容

map.has_key()判断字典中是否存在某键

map[n]将键n的值设为**