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Copy pathContinuous Subarray Sum.java
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101 lines (86 loc) · 2.57 KB
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1522910259
tags: Math, DP, Coordinate DP, Subarray
给一个非负数的数列和数字k(可正负, 可为0). 找到连续子序列(长度超过2), 使得这个subarray的sum 是 k的倍数. 问: 是否可能?
#### DP
- O(n^2)
- 需要记录在0 ~ i点(包括nums[i], 以nums[i]结尾)的sum, 坐标型动态规划.
- dp[i] = dp[i - 1] + nums[i];
- 最后移动, 作比较
#### 直接算结果
- 从sum = 每次[i ~ j]的所有情况
- 验证
```
/*
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.
*/
/*
Thoughts:
If iterate over range[0 ~ n], the move the range by [1 ~ n] steps
Time will become O(n^2)
Utilize rangeSum[0 ~ i] and rangeSum[0 ~ j]
Calculate the diff
dp[i] = sum up to index i
dp[0]: nums[0]
dp[i] = dp[i - 1] + nums[i];
O(n^2)
*/
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return false;
}
int n = nums.length;
int[] dp = new int[n]; // sum up to index i
dp[0] = nums[0];
for (int i = 1; i < n; i++) {
dp[i] = dp[i - 1] + nums[i];
}
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int sum = dp[j] - dp[i] + nums[i];
if (sum == k || (k != 0 && sum % k == 0)) {
return true;
}
}
}
return false;
}
}
/*
Thoughts:
If iterate over range[0 ~ n], the move the range by [1 ~ n] steps
Time will become O(n^2)
*/
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return false;
}
int n = nums.length;
for (int i = 0; i < n; i++) {
int sum = nums[i];
for (int j = i + 1; j < n; j++) {
sum += nums[j];
if (sum == k || (k != 0 && sum % k == 0)) {
return true;
}
}
}
return false;
}
}
```