【LeetCode】560.Subarray Sum Equals K解题报告

【LeetCode】560.Subarray Sum Equals K解题报告

tags: Array

题目地址：https://leetcode.com/problems/subarray-sum-equals-k/tabs/description
题目描述：

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

Examples:

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].

Solutions:

解法一：暴力解法，三层循环，算出所有连续组合的和，为k时count加一，显然超时。时间复杂度O(n³),空间复杂度O(1)。

public class Solution {    public int subarraySum(int[] nums, int k) {        int count = 0;        for (int start = 0; start < nums.length; start++) {            for (int end = start + 1; end <= nums.length; end++) {                int sum = 0;                for (int i = start; i < end; i++)                    sum += nums[i];                if (sum == k)                    count++;            }        }        return count;    }}

解法二：通过循环计算出从第一个数到每一个数的和，那么任意一段连续数字的和就可以通过两段和相减得出。注意好下标以及长度的相关问题。时间复杂度O(n²),空间复杂度O(n)。

public class Solution {    public int subarraySum(int[] nums, int k) {        int count = 0;        int[] sum = new int[nums.length + 1];        sum[0] = 0;        for (int i = 1; i <= nums.length; i++)            sum[i] = sum[i - 1] + nums[i - 1];        for (int start = 0; start < nums.length; start++) {            for (int end = start + 1; end <= nums.length; end++) {                if (sum[end] - sum[start] == k)                    count++;            }        }        return count;    }}

解法三：此法和法一相似，但是此法很巧，在于减少一层循环，每一循环sum要清零。时间复杂度O(n²),空间复杂度O(1)。

public class Solution {    public int subarraySum(int[] nums, int k) {        int count = 0;        for (int start = 0; start < nums.length; start++) {            int sum=0;            for (int end = start; end < nums.length; end++) {                sum+=nums[end];                if (sum == k)                    count++;            }        }        return count;    }}

解法四：使用hashmap。时间复杂度O(n),空间复杂度O(n)。

public class Solution {    public int subarraySum(int[] nums, int k) {        int count = 0, sum = 0;        HashMap < Integer, Integer > map = new HashMap < > ();        map.put(0, 1);        for (int i = 0; i < nums.length; i++) {            sum += nums[i];            if (map.containsKey(sum - k))                count += map.get(sum - k);            map.put(sum, map.getOrDefault(sum, 0) + 1);        }        return count;    }}

Date：2017年7月29日