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:
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<Integer,Integer> map = new HashMap<Integer,Integer>(); 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; } }