Continuous Subarray Sum

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:

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.

思路：

因为(sum + n*k )%k = sum%k,所以，用sum[i]表示从下标0到下标i的累加求和（闭区间）如果sum[i]%k = sum[j] %k,那么(i,j]之间元素求和应该等于n * k,即为k的倍数，所以只需要在判断(i,j]之间元素个数是否满足题意即可。

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {        
        if(nums.length <= 1) return false;
        Map map = new HashMap();
        map.put(0,-1);//防止[0,0],k = 0这种情况
        //(sum + n * k ) % k = ( sum ) % k 
        int sum = 0;
        for(int i = 0; i < nums.length ; i++ ){
            sum += nums[i];
            if( k != 0 )  sum = sum % k;           
            if(map.get(sum) != null){
                int pre = map.get(sum);
                if(i - pre > 1){//因为要求最少是连续两个数之和，如果i与pre相邻的话，说明只有一个数字，而非连续两个
                    return true;
                }
            }else{
                map.put(sum,i);
            }
        }
        return false;
    } 
}