算法第一周Tow Sum[easy]

1.Tow Sum [easy]

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Description

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].

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Solution

此题是在给定数组中找到两个数使其和为target，返回这个数的坐标；
这个题目主要有两个注意的点：

• 数组中的数不能重复使用
• 题目中假定只含有一组解，那我们只需返回拿到的第一满足题目要求的即可

本题采用C语言。

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

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Discussion

本题较为简单我所采用的就是最基本方法，其时间复杂度为O(n2);
通过研究leetcode所给出的answer，通过使用Hash Table 可以将时间复杂度降为O(n)；