[leetcode]523. Continuous Subarray Sum连续子数组和(为K的倍数)
Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.
Example 1:
Input: [23, 2, 4, 6, 7], k=6 Output: True Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
Example 2:
Input: [23, 2, 6, 4, 7], k=6 Output: True Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.
Note:
- The length of the array won't exceed 10,000.
- You may assume the sum of all the numbers is in the range of a signed 32-bit integer.
题目
给定数组和一个数K，求是否存在子数组和为K的倍数。
思路
代码
1 class Solution { 2 public boolean checkSubarraySum(int[] nums, int k) { 3 HashMap<Integer, Integer> map = new HashMap<>(); 4 //为何 map.put(0, -1) 呢？ 如果在第2位找到了mod == 0的数，那就 1 -（-1）>1，return true。 5 map.put(0, -1); 6 int sum = 0; 7 for (int i = 0; i < nums.length; i++) { 8 // running sum 9 sum += nums[i]; 10 if (k != 0) { 11 sum %= k; 12 } 13 // find sum % k is in the HashMap 14 if (map.containsKey(sum)) { 15 // subarray length at least two 16 if (i - map.get(sum) > 1) { 17 return true; 18 } 19 } 20 else { 21 // key: runnng sum -- value: index 22 map.put(sum, i); 23 } 24 } 25 return false; 26 } 27 }