算法分析与设计——LeetCode Problem.1 Two Sum

问题详情

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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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问题分析及思路

刚开始我以为此题可以进行排序后，从后向前直接排除一些大于target的数，然后从前往后再遍历对两个数之和进行判断，若大于target则从后向前一位，再从开始遍历。直到找到等于target之和的两数。
但后来发现这个方法太复杂，此题较为简单不需要这样做，不排序直接从前从后遍历，保证从后遍历的i大于从前遍历的j，同时两者之和等于target时返回即可。

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具体代码

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> des;
        int i = nums.size();
        i--;
        int j = 0;
        for(i; i > j; i--) {
            for(j; j < i; j++) {
                if(nums[j] + nums[i] == target) {
                    des.push_back(j);
                    des.push_back(i);
                    return des;
                }
            }
            j = 0;
        }
    }
};