leetcode560. Subarray Sum Equals K

560. Subarray Sum Equals K

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].

解法一

暴力法，两层循环，判断以每一个元素开头一直累加是否能和k相等。

public class Solution {
    public int subarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int sum  = 0, count = 0;
        for (int i = 0; i < nums.length; i++) {
            sum = 0;
            sum += nums[i];
            if (sum == k) {
                count++;
            }
            for (int j = i + 1; j < nums.length; j++) {
                sum += nums[j];
                if (sum == k) {
                    count++;
                }
            }
        }

        return count;
    }
}

解法二

遍历数组，累加到每一个元素的sum添加到map中，得到0, sum0，sum1，sum2, …., sum(n-1).如果sumj - sumi = k，即可得sum(i+1…j)为k，即计数+1。

public class Solution {
    public int subarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int sum = 0, result = 0;
        Map<Integer, Integer> preSum = new HashMap<>();
        preSum.put(0, 1);

        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (preSum.containsKey(sum - k)) {
                result += preSum.get(sum - k);
            }
            if (preSum.containsKey(sum)) {
                preSum.put(sum, preSum.get(sum) + 1);
            } else {
                preSum.put(sum, 1);
            }
        }

        return result;
    }
}