LeetCode（Two Sum）

一、题目要求

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].

二、解法

C语言

分析：采用的两次for循环进行遍历，算法复杂度O(n^2)

 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     for(int i = 0; i < numsSize - 1; i++) {
 7         int a = nums[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                 return result;
13             }
14         }
15     }
16     return result;
17 }

运行结果：

Python

分析：利用dict结构的查找特性

 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         if len(nums) < 2:
 9             return False
10         
11         buff_dict = {}
12         for i in range(len(nums)):
13             if nums[i] in buff_dict:
14                 return [buff_dict[nums[i]], i]
15             else :
16                 buff_dict[target - nums[i]] = i

运行结果

LeetCode（Two Sum）

一、题目要求

二、解法

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