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Add solution for 0523.Continuous Subarray Sum

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	`1`	`+# Continuous Subarray Sum`
	`2`	`+`
	`3`	`+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 a multiple of k, that is, sums up to n*k where n is also an integer.`
	`4`	`+`
	`5`	`+## Example 1:`
	`6`	+```
	`7`	`+ Input: [23, 2, 4, 6, 7], k=6`
	`8`	`+ Output: True`
	`9`	`+ Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.`
	`10`	+```
	`11`	`+`
	`12`	`+## Example 2:`
	`13`	+```
	`14`	`+ Input: [23, 2, 6, 4, 7], k=6`
	`15`	`+ Output: True`
	`16`	`+ Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.`
	`17`	+```
	`18`	`+`
	`19`	`+`
	`20`	`+`
	`21`	`+Note:`
	`22`	`+`
	`23`	`+ 1. The length of the array won't exceed 10,000.`
	`24`	`+ 2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.`

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	`1`	`+class Solution {`
	`2`	`+ public boolean checkSubarraySum(int[] nums, int k) {`
	`3`	`+ for (int start = 0; start < nums.length; start++) {`
	`4`	`+ int check = 0;`
	`5`	`+ for (int i = start; i < nums.length; i++) {`
	`6`	`+ check += nums[i];`
	`7`	`+ if (i > start) {`
	`8`	`+ if (k != 0) {`
	`9`	`+ if (check % k == 0) {`
	`10`	`+ return true;`
	`11`	`+ }`
	`12`	`+ } else {`
	`13`	`+ if (check == k) {`
	`14`	`+ return true;`
	`15`	`+ }`
	`16`	`+ }`
	`17`	`+ }`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+`
	`21`	`+ return false;`
	`22`	`+ }`
	`23`	`+}`

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