leetcode 523. Continuous Subarray Sum

标签：思路   ash   order   ast   range   bool   bar   get   array   

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.

Example 1:

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

Example 2:

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

Note:

1. The length of the array won‘t exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

思路

(sum[x] - sum[y])%k == 0  等同于  sum[x] %k - sum[y]%k == 0  等同判断  sum[x]%k  和 sum[y]%k 是否相等

最恶心的是 这到题目 k 可能为0  而且要求 长度大于等于2

用 hash mp记录这个数存在的同时还要记录他的位置信息，这样在判断长度的时候就方便很多了。

class Solution {  
public:  
    bool checkSubarraySum(vector<int>& nums, int k) {  
        unordered_map<int, int> mp;  
        int sum = 0;  
        mp[0] = -1;  
        for(int i = 0; i < nums.size(); ++i) {  
            sum += nums[i];  
            if (k) sum %= k;  
            if (mp.count(sum) > 0) {  
                if (i - mp[sum] > 1) return true;  
            }  
            else mp[sum] = i;  
        }
        return false;  
    }  
}; 

leetcode 523. Continuous Subarray Sum

标签：思路   ash   order   ast   range   bool   bar   get   array   

原文地址：http://www.cnblogs.com/pk28/p/7454546.html