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.思路：
We iterate through the input array exactly once, keeping track of the running sum mod k of the elements in the process. If we find that a running sum value at index j has been previously seen before in some earlier index i in the array, then we know that the sub-array (i,j] contains a desired sum.
乍一看没看懂上面的英语，其实它就是这个意思： 对一个数组【a1,a2,a3,a4,a5…am…an…】，用A表示数组每一个元素的累加和 Am = a1 +a2 +a3 +…am An = a1 +a2 +a3 +…an 如果 Am % k == An % k ，则(An-Am) % k= 0。 说明Am到An之间的数字的和，除以k的余数也为0。 那么，这些数字就是我们要找的一个连续的子串。
java 代码如下：
class Solution { public boolean checkSubarraySum(int[] nums, int k) { if(nums.length == 1) return false; if(k == 0) { for(int i = 0; i < nums.length; i++) { if(nums[i] != 0) return false; } return true; } HashMap<Integer, Integer> map = new HashMap(); // 余数 : 索引下标 map.put(0, -1); int sum = 0; for(int i = 0; i < nums.length; i++) { sum += nums[i]; sum %= k; if(map.containsKey(sum)) { if(i - map.get(sum) > 1) return true; }else { map.put(sum, i); } } return false; } }注：学渣心里苦，不要学楼主，平时不努力，考试二百五，哭 ~
