题目:
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:
Add one piece of fruit from this tree to your baskets. If you cannot, stop.
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 <= tree.length <= 40000
0 <= tree[i] < tree.length
思路:
这道题是非常典型的滑动窗口,虽然leetcode把它定义为medium,但是由于它限制了篮子的数目为2,所以其实比easy难度高不了多少.
我们首先从第一颗树开始收果子,然后如果收到的果子种类types>2,那么说明要抛弃前面的同种类的果子知道types<=2,然后继续往下收果子就行,可以用一个数组来表示收到的果子种类types
代码:
class Solution {
public int totalFruit(int[] tree) {
int [] arr = new int [40005];
int types=0;
int start=0;
int maxLen=-1;
for(int i=0;i<tree.length;i++){
if(arr[tree[i]]==0)
types++;
arr[tree[i]]++;
while(types>2){
arr[tree[start]]--;
if(arr[tree[start]]==0)
types--;
start++;
}
if(types==2&&maxLen<i-start+1){
maxLen=i-start+1;
}
}
return (types==1)?tree.length:maxLen;
}
}
分析:
可知,第一个for循环和while(type>2)是串行的,也就是说不是每个i都会进入while循环,也不是每个i都会在while中循环i-1次,仔细观察可知虽然有两个嵌套循环,但其实他们复杂度是并列的,所以最终时间复杂度为O(n)
这里只有两个篮子,而如果篮子数目不定,为k,那么这道题的2要改成k,且最后的return部分的==1要改成!=k,详情可以见滑动窗口系列的(三)
