Leetcode 523.Continuous Subarray Sum
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:
Example 2:
Input: [23, 2, 6, 4, 7], k=6
Output: True
Explanation:
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.
乍一看沒看懂上面的英語,其實它就是這個意思:
對一個數組【a1,a2,a3,a4,a5…am…an…】,用A表示數組每一個元素的累加和
Am = a1 +a2 +a3 +…am
An = a1 +a2 +a3 +…an
如果 Am % k == An % k ,則(An-Am) % k= 0。
說明Am到An之間的數字的和,除以k的餘數也為0。
那麼,這些數字就是我們要找的一個連續的子串。
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(); // 餘數 : 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;
}
}