这道题目一开始的思路是利用水果的种类大于等于三，来作为滑动窗口的维护条件，使用两个key值来记录两种水果的值，当遇到第三种水果时，则将slowindex设置为slowindex-1，然后将slowindex逐渐缩小，来查找前x个相同的元素，之后重新设置key值，继续搜索。具体代码如下：

class Solution {
public:int totalFruit(vector<int>& fruits) {int slowindex = 0;int fastindex = 1;int sum= 1;int kind = 1;int key1 = fruits[slowindex];int key2 = -1;if(fruits.size() == 0){return 0;}while(fastindex < fruits.size()){if(kind < 2){if(fruits[fastindex] == key1){fastindex++;sum++;}else if(key2 < 0){key2 = fruits[fastindex];kind++;sum++;fastindex++;}}if(fruits[fastindex] == key1 || fruits[fastindex] == key2){sum++;fastindex++;}else{kind++;}while(kind >= 3){slowindex = fastindex - 1;key1 = fruits[slowindex--];while(fruits[slowindex] == key1){slowindex--;}key2 = fruits[fastindex];kind--;fastindex++;}sum = max(sum, fastindex - slowindex - 1);}return sum;}
};

但很快就出了问题，在测试用例 [3,3,3,1,2,1,1,2,3,3,4]时，预期结果为5，而我的代码输出结果却为8，在思考后得出结论：

• sum的更新和维护，在代码1中，没有用到currentsum来记录当前的最大值，仅仅使用了一个sum来计数，导致这个sum在满足条件 fruits[fastindex] == key1 || fruits[fastindex] == key2 时便会增加，从而导致sum不断增加，fastindex - slowindex - 1 不会被取到。修改后的代码如下：

class Solution {
public:int totalFruit(vector<int>& fruits) {int slowindex = 0;int fastindex = 0;int sum = 0;int currentsum = 0;int key1 = fruits[slowindex];int key2 = -1;while(fastindex < fruits.size()) {if (fruits[fastindex] == key1 || fruits[fastindex] == key2) {currentsum++;fastindex++;} else{if (key2 == -1) {key2 = fruits[fastindex];currentsum++;fastindex++;}else{sum = max(sum, currentsum);currentsum = 2;slowindex = fastindex - 1;key1 = fruits[slowindex];while (slowindex > 0 && fruits[slowindex - 1] == key1) {currentsum++;slowindex--;}key2 = fruits[fastindex];fastindex++;}}}sum = max(sum, currentsum);return sum;}
};

这段代码就能够顺利通过了。