这篇文章主要介绍C++实现LeetCode之最短子数组求和的示例分析,文中介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们一定要看完!
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.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
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:
// 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:
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:
// 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:
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; } };
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