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)

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

运行结果：

Python

分析：利用dict结构的查找特性

 class Solution(object):
     def twoSum(self, nums, target):
         """
         :type nums: List[int]
         :type target: int
         :rtype: List[int]
         """
         if len(nums) < 2:
             return False

         buff_dict = {}
         for i in range(len(nums)):
             if nums[i] in buff_dict:
                 return [buff_dict[nums[i]], i]
             else :
                 buff_dict[target - nums[i]] = i

运行结果