560.Subarray Sum Equals K

Posted 2021-01-15 ovs98

Description

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2
Output: 2

Note:

1. The length of the array is in range [1, 20,000].
2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

一.题目理解

给定一个数组，找出其中连续子序列之和为K的子序列个数。

二.题目解答

首先我们想到的是打表求出累加数组，然后再去遍历求解，这种做法很简答，但是很耗时，代码如下

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        vector<int> sums = nums;
        int cur = 0;


        //打表
        for(int i = 1; i < nums.size(); i++)
            sums[i] = sums[i - 1] + nums[i];

        //查找
        cur = 0;
        for(int i = 0; i < nums.size(); i++) {
            if(sums[i] == k)
                cur++;

            for(int j = i - 1; j >= 0; j--) {
                if(sums[i] - sums[j] == k)
                    cur++;

            }
        }

        return cur;
    }
};

上述代码用了两重循环，因而是很耗时的，我们考虑能否降低循环次数？

想了半天没想到，找到网上一种无序表的方法，无序表建立在连续数组之和与其出现次数之间的映射上。

下图是我的解释

 

代码如下：

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {

       int cur = 0;
       unordered_map<int,int> umap{{0,1}};

       int sum = 0;
       for(int i = 0;i < nums.size();i++){
        sum += nums[i];

        cur += umap[sum - k];
        ++umap[sum];
       }

        return cur;
    }
};