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(2 * sizeof(int));
    for(int i = 0; i < numsSize - 1; i++) {
        int a = nums[i];
        for(int j = i + 1; j < numsSize; j++ ) {
            if(nums[i] + nums[j] == target) {
                result[0] = i;
                result[1] = 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