[LeetCode]167. Two Sum II - Input array is sorted

[LeetCode]167. Two Sum II - Input array is sorted

题目描述

思路

开始的思路是暴力解，虽然用例都通过了，但是超时了
看了超时的用例，发现是太多重复造成的
于是加了重复检测的条件判断，提交通过之后发现速度太慢
换一种思路
两个指针从数组两边同时向中间移动，如果数字之和小于目标，则移动左指针，否则移动右指针

代码

注释部分为暴力解法

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    /*
    vector<int> twoSum(vector<int>& numbers, int target) {
        int current, next;
        vector <int> res;
        bool finded = false;
        for (current = 0; current < numbers.size(); ++ current) {
            next = current + 1;

            finded = false;
            for (; next < numbers.size(); ++next) {
                if (numbers[current] + numbers[next] == target) {
                    finded = true;
                    res.push_back(current + 1);
                    res.push_back(next + 1);
                    break;
                }
            }

            if (finded)
                break;

            if (numbers[current] == numbers[next]) {
                ++current;
                continue;
            }
        }
        return res;
    }
    */
    vector<int> twoSum(vector<int>& numbers, int target) {
        int low = 0, high = numbers.size() - 1;
        vector<int> res;
        while (numbers[low] + numbers[high] != target) {
            if (numbers[low] + numbers[high] < target) {
                ++low;
            }
            else {
                --high;
            }
        }
        res.push_back(low + 1);
        res.push_back(high + 1);

        return res;
    }
};

int main() {
    vector<int> nums = { 0, 0, 3, 4 }, res;
    Solution s;
    res = s.twoSum(nums, 0);
    for (int p : res) {
        cout << p << " ";
    }
    system("pause");
}