标签：return each [1] sam numbers one blog and 申请
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].
1 /** 2 * Note: The returned array must be malloced, assume caller calls free(). 3 */ 4 int* twoSum(int* nums, int numsSize, int target) { 5 int *result = (int*)malloc(2 * sizeof(int)); 6 7 for(int i = 0; i < numsSize-1; i++){ 8 for(int j = i+1; j < numsSize; j++){ 9 if((nums[i] + nums[j]) == target){ 10 result[0] = i; 11 result[1] = j; 12 } 13 } 14 } 15 return result; 16 17 }
注意使用malloc为指针申请空间
1 class Solution(object): 2 def twoSum(self, nums, target): 3 """ 4 :type nums: List[int] 5 :type target: int 6 :rtype: List[int] 7 """ 8 map= {} 9 for i,n in enumerate(nums): 10 if map.has_key(n): 11 return map[n],i 12 else: 13 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**
标签：return each [1] sam numbers one blog and 申请
原文地址：http://www.cnblogs.com/fullest-life/p/6475927.html
