LeetCode560. Subarray Sum Equals K

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:

The length of the array is in range [1, 20,000].

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

my program

思路：创建一个数组tmp，用来储存前n项的和。

第一层循环：用来计算每一个前i项的和，判断如果前i项和等于k，则结果res加一；

内层循环：遍历tmp的前i项，判断如果（前i项和-前j项和）== k，则结果res加一；

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        vector<int> tmp = nums;
        int res = 0;
        if (nums[0] == k) res=1;
        for (int i = 1; i<nums.size(); i++) {
            tmp[i] += tmp[i-1];
            if (tmp[i] == k) res++;
            for (int j = 0; j < i; j++) {
                if (tmp[i] - tmp[j] == k) res++;
            }
        }
        return res;
    }
};

Submission Details

80 / 80 test cases passed.

Status: Accepted

Runtime: 369 ms

虽然AC了，但是runtime很高，于是思考是否有更有的算法。

暴力解法

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int res = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            int sum = nums[i];
            if (sum == k) ++res;
            for (int j = i + 1; j < n; ++j) {
                sum += nums[j];
                if (sum == k) ++res;
            }
        }
        return res;
    }
};

Submission Details

80 / 80 test cases passed.

Status: Accepted

Runtime: 538 ms

两种解法的时间复杂度均为O(n2),类似

哈希表解法

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int res = 0;
        map<int, int> tmp{{0,1}};
        int sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
            res += tmp[sum - k];
            ++tmp[sum];
        }
        return res;
    }
};

非常巧妙，时间复杂度仅为O(n),相当于仅仅遍历了数组。