题目要求

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 a 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:

1. The length of the array won't exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

假设有一个非负整数数组，要求找到其中一个连续的子数组，该子数组中所有元素的和是k的整数倍。要求连续子数组的长度大于等于2.

思路一：暴力循环

如果我们可以计算出所有长度大于等于2的连续子数组的元素和，就可以判断出是否存在满足题意的子数组。这里可以做一个简单的优化，即用一个额外的数组用来存[0..i]的子数组的元素和，则sum[i..j]的子数组的元素和可以通过sum[0..j+1]-sum[0..i-1]计算得出。代码如下：

    public boolean checkSubarraySum(int[] nums, int k) {  
    if (nums == null || nums.length <= 1) {  
        return false;  
    }  
    int[] sum = new int[nums.length+1];  
    sum[0] = 0;  
    for (int i = 1 ; i<sum.length ; i++) {  
        sum[i] += sum[i-1] + nums[i-1];  
    }  
    for (int i = nums.length ; i>=2 ; i--) {  
        for (int j = 0 ; j+i<sum.length ; j++) {  
            int diff = sum[j+i] - sum[j];  
            if (diff == k || (k!=0 && diff % k == 0)) {  
                return true;  
            }  
        }  
    }  
    return false;  
}

思路二：记录中间结果

基于思路一的基础上可以进一步优化，既然我们要计算的是k的整数倍的子数组和，而我们有的是可以通过O(N)的时间复杂度计算出[0...i]子数组的元素和。那么我们可以推算出，假如sum[0..j]%k = m，而sum[0..i]%k = m(i-j>=2)，那么可以知道[i+1, j]这个子数组一定是k的整数倍。因此我们只需要记录已经遍历过的从0下标开始的所有子数组的和以及对应的下标值，并且判断是否存在如上述的关联即可。

public boolean checkSubarraySum2(int nums[], int k){  
  
    Map<Integer,Integer> map = new HashMap<>();  
    map.put(0,-1);  
    int sumSoFar = 0;  
    for(int i=0; i < nums.length; i++){  
        sumSoFar = sumSoFar + nums[i];  
        if(k != 0) sumSoFar = sumSoFar % k;  
        if(map.containsKey(sumSoFar)){  
            int start = map.get(sumSoFar);  
            if(i-start > 1) return true;  
        }else{  
            map.put(sumSoFar, i);  
        }  
    }  
    return false;  
}

思路三：分治法

分治法在这题的核心思想在于，将整个数组先一分为二，分别判断在左子数组和右子数组中是否存在满足条件的子数组，如果没有，再判断跨左右子数组的子数组是否存在满足条件的连续子数组。分治法和动态规划都在于想通递归场景后，代码就行云流水了。

public boolean checkSubarraySum(int[] nums, int lo, int hi, int k){  
    if(lo==hi) return false;  
    int mid = lo+(hi-lo)/2;  
    if(checkSubarraySum(nums, lo, mid, k)) return true;  
    if(checkSubarraySum(nums, mid+1, hi, k)) return true;  
    int left = mid, right = mid+1;  
    int sum = nums[left];  
    while(left>=lo&&right<=hi){  
        sum += nums[right];  
        if((k>0&&sum%k==0)||(k==0&&sum==0)) return true;  
        --left;  
        ++right;  
        if(left>=lo){  
            sum += nums[left];  
            if((k>0&&sum%k==0)||(k==0&&sum==0))  return true;  
        }  
    }  
    return false;  
}  
public boolean checkSubarraySum(int[] nums, int k) {  
    k = Math.abs(k);  
    return checkSubarraySum(nums, 0, nums.length-1, k);  
}