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560.Subarray-Sum-Equals-K

560. Subarray Sum Equals K

题目地址

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.
​
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].

代码

Approach #2 Commulative Sum

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] - start[start] == k) {
count++;
}
}
}
​
return count;
}
}

Approach #3 Without space

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;
}
}

Approach #4 Hashmap

Complexity Analysis
  • Time complexity : O(n)
  • Space complexity : O(n)
class Solution {
public int subarraySum(int[] nums, int k) {
int count = 0;
int 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;
}
}

Approach #1 Brute Force [Time Limit Exceeded]

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;
}
}