思路：先把字符串换成数组，用01234代表aeiou，方便比大小。用last记录上一个数字，当前扫描的数字为arr[i]，合法的情况只有arr[i] == last 或者arr[i] == last+1。否则，就得重新开始找合法的子串。

在更新max_len的时候，只有所有元音字母都包含才行(checkALL())。重新找合法子串时，记得将记录是否包含的标记(known[])。但这里有一个小插曲，我起初是使用.resize(5, false)来清空的，后来发现这样子不会达到预想的效果，很多标记还没有被置为false，于是重新写了个函数来完清空。

        for (int i = 0; i < arr.size(); i++) {if (arr[i] == last + 1) {known[arr[i]] = true;tmp_len++;if (checkAll())max_len = max(max_len, tmp_len);}else if (arr[i] == last){tmp_len++;if (checkAll())max_len = max(max_len, tmp_len);}else {tmp_len = 1;doClear();known[arr[i]] = true;}last = arr[i];}

完整代码

class Solution {
public:vector known;vector arr;bool checkAll() {for (int i  = 0; i < known.size(); i++) {if (!known[i])return false;}return true;}void doClear() {for (int i  = 0; i < known.size(); i++)known[i] = false;}int longestBeautifulSubstring(string word) {arr.clear();for (int i = 0; i < word.length(); i++) {char c = word[i];switch(word[i]) {case 'a': arr.push_back(0);break;case 'e': arr.push_back(1);break;case 'i': arr.push_back(2);break;case 'o': arr.push_back(3);break;default: arr.push_back(4);}}known.resize(5, false);int last = -1;int max_len = 0, tmp_len = 0; for (int i = 0; i < arr.size(); i++) {if (arr[i] == last + 1) {known[arr[i]] = true;tmp_len++;if (checkAll())max_len = max(max_len, tmp_len);}else if (arr[i] == last){tmp_len++;if (checkAll())max_len = max(max_len, tmp_len);}else {tmp_len = 1;doClear();known[arr[i]] = true;}last = arr[i];}if (max_len < 5)max_len = 0;return max_len;        }
};