leetcode209--长度最小的子数组

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;
                else
                    r = m;

            }
           // printf("%d,%d\n", i, l);
            if ( l < sz){
                ans = min(l - i + 1, ans);
            }
        }


        return ans == INT_MAX ? 0 : ans;
    }
};