description:
523 | Continuous Subarray Sum | 23.2% | Medium |
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.
thought:
这道题没想出什么创新的方法,所以就不想po自己的代码上去了,看了一下别人的solution,发现可以先把和求出来,然后再相减的方法,进行连续子串和的计算,然后取模,感觉是一种比较巧妙地方法了。
Better ways:
by tankztc
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
if (nums == null || nums.length == 0) return false;
int[] preSum = new int[nums.length+1];
for (int i = 1; i <= nums.length; i++) {
preSum[i] = preSum[i-1] + nums[i-1];
}
for (int i = 0; i < nums.length; i++) {
for (int j = i+2; j <= nums.length; j++) {
if (k == 0) {
if (preSum[j] - preSum[i] == 0) {
return true;
}
} else if ((preSum[j] - preSum[i]) % k == 0) {
return true;
}
}
}
return false;
}
};