[LeetCode] 209. Minimum Size Subarray Sum 最短子数组之和
Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
Example:
Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
这道题给定了我们一个数字,让求子数组之和大于等于给定值的最小长度,注意这里是大于等于,不是等于。跟之前那道 Maximum Subarray 有些类似,并且题目中要求实现 O(n) 和 O(nlgn) 两种解法,那么先来看 O(n) 的解法,需要定义两个指针 left 和 right,分别记录子数组的左右的边界位置,然后让 right 向右移,直到子数组和大于等于给定值或者 right 达到数组末尾,此时更新最短距离,并且将 left 像右移一位,然后再 sum 中减去移去的值,然后重复上面的步骤,直到 right 到达末尾,且 left 到达临界位置,即要么到达边界,要么再往右移动,和就会小于给定值。代码如下:
解法一:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
|
// O(n)
class Solution {
public :
int minSubArrayLen( int s, vector< int >& nums) {
if (nums.empty()) return 0;
int left = 0, right = 0, sum = 0, len = nums.size(), res = len + 1;
while (right < len) {
while (sum < s && right < len) {
sum += nums[right++];
}
while (sum >= s) {
res = min(res, right - left);
sum -= nums[left++];
}
}
return res == len + 1 ? 0 : res;
}
};
|
同样的思路,我们也可以换一种写法,参考代码如下:
解法二:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
|
class Solution {
public :
int minSubArrayLen( int s, vector< int >& nums) {
int res = INT_MAX, left = 0, sum = 0;
for ( int i = 0; i < nums.size(); ++i) {
sum += nums[i];
while (left <= i && sum >= s) {
res = min(res, i - left + 1);
sum -= nums[left++];
}
}
return res == INT_MAX ? 0 : res;
}
};
|
下面再来看看 O(nlgn) 的解法,这个解法要用到二分查找法,思路是,建立一个比原数组长一位的 sums 数组,其中 sums[i] 表示 nums 数组中 [0, i - 1] 的和,然后对于 sums 中每一个值 sums[i],用二分查找法找到子数组的右边界位置,使该子数组之和大于 sums[i] + s,然后更新最短长度的距离即可。代码如下:
解法三:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
// O(nlgn)
class Solution {
public :
int minSubArrayLen( int s, vector< int >& nums) {
int len = nums.size(), sums[len + 1] = {0}, res = len + 1;
for ( int i = 1; i < len + 1; ++i) sums[i] = sums[i - 1] + nums[i - 1];
for ( int i = 0; i < len + 1; ++i) {
int right = searchRight(i + 1, len, sums[i] + s, sums);
if (right == len + 1) break ;
if (res > right - i) res = right - i;
}
return res == len + 1 ? 0 : res;
}
int searchRight( int left, int right, int key, int sums[]) {
while (left <= right) {
int mid = (left + right) / 2;
if (sums[mid] >= key) right = mid - 1;
else left = mid + 1;
}
return left;
}
};
|
我们也可以不用为二分查找法专门写一个函数,直接嵌套在 for 循环中即可,参加代码如下:
解法四:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
|
class Solution {
public :
int minSubArrayLen( int s, vector< int >& nums) {
int res = INT_MAX, n = nums.size();
vector< int > sums(n + 1, 0);
for ( int i = 1; i < n + 1; ++i) sums[i] = sums[i - 1] + nums[i - 1];
for ( int i = 0; i < n; ++i) {
int left = i + 1, right = n, t = sums[i] + s;
while (left <= right) {
int mid = left + (right - left) / 2;
if (sums[mid] < t) left = mid + 1;
else right = mid - 1;
}
if (left == n + 1) break ;
res = min(res, left - i);
}
return res == INT_MAX ? 0 : res;
}
};
|
讨论:本题有一个很好的 Follow up,就是去掉所有数字是正数的限制条件,而去掉这个条件会使得累加数组不一定会是递增的了,那么就不能使用二分法,同时双指针的方法也会失效,只能另辟蹊径了。其实博主觉得同时应该去掉大于s的条件,只保留 sum=s 这个要求,因为这样就可以在建立累加数组后用 2sum 的思路,快速查找 s-sum 是否存在,如果有了大于的条件,还得继续遍历所有大于 s-sum 的值,效率提高不了多少。
Github 同步地址:
https://github.com/grandyang/leetcode/issues/209
类似题目:
Minimum Window Substring
参考资料:
https://leetcode.com/problems/minimum-size-subarray-sum/
https://leetcode.com/problems/minimum-size-subarray-sum/discuss/59090/C%2B%2B-O(n)-and-O(nlogn)
https://leetcode.com/problems/minimum-size-subarray-sum/discuss/59078/Accepted-clean-Java-O(n)-solution-(two-pointers)
到此这篇关于C++实现LeetCode(209.最短子数组之和)的文章就介绍到这了,更多相关C++实现最短子数组之和内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家!
原文链接:https://www.cnblogs.com/grandyang/p/4501934.html