https://leetcode.com/problems/fruit-into-baskets/

In a row of trees, the i-th tree produces fruit with type tree[i].

You start at any tree of your choice, then repeatedly perform the following steps:

  1. Add one piece of fruit from this tree to your baskets.  If you cannot, stop.
  2. Move to the next tree to the right of the current tree.  If there is no tree to the right, stop.

Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop.

You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each.

What is the total amount of fruit you can collect with this procedure?

 

Example 1:

Input: [1,2,1]
Output: 3
Explanation: We can collect [1,2,1].

Example 2:

Input: [0,1,2,2]
Output: 3
Explanation: We can collect [1,2,2].
If we started at the first tree, we would only collect [0, 1].

Example 3:

Input: [1,2,3,2,2]
Output: 4
Explanation: We can collect [2,3,2,2].
If we started at the first tree, we would only collect [1, 2].

Example 4:

Input: [3,3,3,1,2,1,1,2,3,3,4]
Output: 5
Explanation: We can collect [1,2,1,1,2].
If we started at the first tree or the eighth tree, we would only collect 4 fruits.

 

Note:

  1. 1 <= tree.length <= 40000
  2. 0 <= tree[i] < tree.length

开始理解错题意，没考虑到1，1，2，1，1，3这种1的不连续

调试两天也没出来

还是看了别人代码才写的

p.s. map是基于红黑树 unordered_map基于哈希，所以前者慢。

思路就是把分段的记录放入map，滑动窗口的思想，维护map元素大小小于2

class Solution {
public:
    int totalFruit(vector<int>& tree) {
        int st = 0;
        unordered_map<int, int>m;
        int maxn = 0;
        for (int i = 0; i < tree.size(); i ++) {
            m[tree[i]] ++;
            while (m.size() > 2) {
                m[tree[st]] --;
                if (m[tree[st]] == 0)
                    m.erase(tree[st]);
                st ++;
            }
            maxn = max(maxn, i - st + 1);
        }
        return maxn;
    }
};