Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.
For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.
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，让我们求子数组之和大于等于给定值的最小长度，跟之前那道 Maximum Subarray 最大子数组有些类似，并且题目中要求我们实现O(n)和O(nlgn)两种解法，那么我们先来看O(n)的解法，我们需要定义两个指针left和right，分别记录子数组的左右的边界位置，然后我们让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，此时我们更新最短距离，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; } };
下面我们再来看看O(nlgn)的解法，这个解法要用到二分查找法，思路是，我们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，然后我们对于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，然后我们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; } };
本文转自博客园Grandyang的博客，原文链接：最短子数组之和[LeetCode] Minimum Size Subarray Sum ，如需转载请自行联系原博主。
