[LintCode] 1844. subarray sum equals k II

Given an array of integers and an integer k, you need to find the minimum size of continuous subarrays whose sum equals to k, and return its length.

if there are no such subarray, return -1.

Example

Example1

Input: nums = [1,1,1,2] and k = 3
Output: 2

Example2

Input: nums = [2,1,-1,4,2,-3] and k = 3
Output: 2

Notice

the integer nums[i] may lower than 0

Solution 1: O(N^2) TLE

public class Solution {
    /**
     * @param nums: a list of integer
     * @param k: an integer
     * @return: return an integer, denote the minimum length of continuous subarrays whose sum equals to k
     */
    public int subarraySumEqualsKII(int[] nums, int k) {
        // write your code here
        int[] prefix = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (i == 0) {
                prefix[i] = nums[i];
                continue;
            }
            prefix[i] = prefix[i - 1] + nums[i];
        }
        System.out.println(Arrays.toString(prefix));

        int res = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i; j < nums.length; j++) {
                if (i == 0) {
                    if (prefix[j] == k) {
                        res = Math.min(res, j + 1);
                    }
                    continue;
                }
                if (prefix[j] - prefix[i - 1] == k) {
                    res = Math.min(res, j - i + 1);
                }
            }
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

Solution 2:

public class Solution {
    /**
     * @param nums: a list of integer
     * @param k: an integer
     * @return: return an integer, denote the minimum length of continuous subarrays whose sum equals to k
     */
    public int subarraySumEqualsKII(int[] nums, int k) {
        // write your code here
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        int res = Integer.MAX_VALUE;
        map.put(0, -1);

        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (map.containsKey(sum - k)) {
                // actually get the i - 1 index for  map.get(sum - k)
                res = Math.min(res, i - map.get(sum - k));
            }
            map.put(sum, i);
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

原文地址：https://www.cnblogs.com/xuanlu/p/12515155.html