523. 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=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: 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.

題意:

給定一個非負的數組，還有一個目標數k，寫一個方法來找到這個數組是否有連續的子數組(至少兩個)可以加總為k的倍數。

題解:

需要考慮兩種情

*一般情況*

從每個數組的元素逐一拜訪，然後每一次循環的過程中再逐一拜訪該元素之後的所有元素，檢查是否加總為k的倍數。

*k為0的情況*

若k=0，問題就變成是否有子數組有連續兩個0的情況。

package LeetCode.Medium;public class ContinuousSubarraySum {    public boolean checkSubarraySum(int[] nums, int k) {                //考慮k為0的情況        if(k == 0) {            for(int i = 0; i < nums.length - 1; i ++) {                if(nums[i] == 0 && nums[i + 1] == 0)                    return true;            }            return false;        }                //一般情況        for(int i = 0; i < nums.length; i ++) {            int sub_sum = nums[i];            for(int j = i + 1; j < nums.length; j ++) {                sub_sum += nums[j];                //子序列為加總為k或是為k的倍數，則為true                if(sub_sum == k || sub_sum % k == 0)                    return true;            }        }                return false;    }}