算法分析与设计课程10——523. Continuous Subarray Sum

一、题目————原题链接

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example1

Input: [23, 2, 4, 6, 7],  k=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example2

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

二、分析

考虑这样一个事实，sum = 0，遍历一个数组[a1, a2, a3, a4, a5]，每次遍历都执行：sum = (ai + sum) % k，并且记录每次的sum值，当sum之前出现过的值再次出现时，我们找到了一个 continuous subarray，因为：令之前sum值相同的数组中位置为j，当前位置为k，(j,k]就是当前k的sum去除之前j的sum后得到的结果——0，即能被k整除。

三、源代码

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        map<int, int> m;        m[0] = -1;        int sum = 0;        for(int i = 0; i < nums.size(); i ++)        {            sum += nums[i];            if(k != 0)            sum %= k;             pair<map<int, int>::iterator, bool> flag = m.insert(pair<int, int>(sum, i));            if(!flag.second)                if(i - m[sum] > 1)                    return true;        }        return false;    }};