209. 长度最小的子数组

https://leetcode.cn/problems/minimum-size-subarray-sum/?envType=study-plan-v2&envId=top-interview-150

思路

三种方法中，具有最优时间复杂度的方案

滑动窗口

设置 start=0 end=0

执行循环：

　　end向后探测，直到sum值大于等于target，更新最优长度

　　然后start向前探测，探测过程中，如果sum仍然大于等于target，更新最优长度，

　　　　start向前探测，直到sum小于target时候停止。

此过程很像毛毛虫先前蠕动行为， 可以称之为毛毛虫行走法。

https://leetcode.cn/problems/minimum-size-subarray-sum/solutions/305704/chang-du-zui-xiao-de-zi-shu-zu-by-leetcode-solutio/?envType=study-plan-v2&envId=top-interview-150

Code

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        /*
        [start, end]
        以此为窗口探测总和>=target的最小连续序列长度
        */
        int start = 0;
        int end = 0;
        long long sum = nums[0];
        int minlen = 1e5+5;

        while(start<=end && end<nums.size()){
            // cout << "------------------------------------------" << endl;

            // cout << "BEGIN ====> start = " << start << " end=" << end << endl;

            // if (sum >= target){
            //     minlen = min(minlen, end-start+1);

            //     cout << "update1 minlen = " << minlen << endl;
            // } 

            do {
                if (sum >= target){
                    minlen = min(minlen, end-start+1);

                    // cout << "update2 minlen = " << minlen << endl;
                    break;
                }

                end++;
                if (end >= nums.size()){
                    break;
                }

                sum += nums[end];
            } while(true);

            if (end >= nums.size()){
                break;
            }

            // cout << "AFTER ENLARGE end ====> start = " << start << " end=" << end << endl;

            do {
                if (sum < target){
                    break;
                } else if (sum >= target){
                    minlen = min(minlen, end-start+1);

                    // cout << "update1 minlen = " << minlen << endl;
                } 

                if (start<end){
                    sum -= nums[start];
                    start++;
                } else {
                    break;
                }
            } while(true);

            // cout << "AFTER ENLARGE start ====> start = " << start << " end=" << end << endl;

            // cout << "minlen = " << minlen << endl;
            // cout << "END ====> start = " << start << " end=" << end << endl;

            if (minlen == 1){
                break;
            }
        }

        return minlen==1e5+5?0:minlen;
    }
};