1. 题意

找到最小满足和大于$target$的子数组长度。
长度最小的子数组

2. 题解

2.1 滑动窗口

class Solution {
public:int minSubArrayLen(int target, vector<int>& nums) {int l = 0;int sum = 0;int sz = nums.size();int ans = INT_MAX;for (int i = 0; i < sz; ++i) {sum += nums[i];if ( sum >= target) {while ( l <= i && sum >= target) {sum -= nums[l];++l;}ans = min(ans, i - l + 2);}}return ans == INT_MAX ? 0 : ans;}
};

2.2 二分

枚举左端点，二分查找满足条件的右端点。

class Solution {
public:bool check(int l, int r, vector<int> &nums, int target){int sum = 0;for (int i = l; i < r + 1; ++i)sum += nums[i];return sum >= target;}int minSubArrayLen(int target, vector<int>& nums) {int l = 0;int sum = 0;int sz = nums.size();int ans = INT_MAX;vector<int> pre(sz + 1, 0);int preSum = 0;for (int i = 0;i < sz; ++i) {preSum += nums[i];pre[i + 1] = preSum;}for (int i = 0;i < sz; ++i) {int l = i;int r = sz;while (l < r) {int m = ((r - l) >> 1) + l;// cout << "m"<< m << "," << "sum" << sum << endl; if ( pre[m + 1] - pre[i] < target)l = m + 1;elser = m;}// printf("%d,%d\n", i, l);if ( l < sz){ans = min(l - i + 1, ans);}}return ans == INT_MAX ? 0 : ans;}
};