523. Continuous Subarray Sum

523. Continuous Subarray Sum

• 題目描述：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.

• 題目大意：給定一個數組，一個k,找出數組中是否存在長度大於等於2的連續的子數組，使連續子數組的和是k的倍數。

• 思路：建立一個runningSum變量，記錄遍歷過程中連續子數組和模k的結果，並將結果存儲到Hashmap中，如果發現當前下標爲j的runningSum,在前面已經出現，則代表在之前出現的位置i到當前位置j，（i,j]處，存在連續子數組和可以整除k。

• 代碼

  package DP;
  
  import java.util.HashMap;
  import java.util.Map;
  
  /**
  * @author OovEver
  * 2018/1/3 10:34
  */
  public class LeetCode523 {
    public static boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int runningSum = 0;
       for(int i=0;i<nums.length;i++) {
            runningSum += nums[i];
            if (k != 0) {
                runningSum %= k;
            }
            Integer prev = map.get(runningSum);
           if (prev != null) {
  //               連續子數組的長度需要大於2
               if (i - prev > 1) {
                   return true;
               }
           } else {
               map.put(runningSum, i);
           }
        }
        return false;
    }
  }