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English Version

题目描述

给你一个字符串 s ,请你返回满足以下条件的最长子字符串的长度:每个元音字母,即 'a','e','i','o','u' ,在子字符串中都恰好出现了偶数次。

 

示例 1:

输入:s = "eleetminicoworoep"
输出:13
解释:最长子字符串是 "leetminicowor" ,它包含 e,i,o 各 2 个,以及 0 个 au 

示例 2:

输入:s = "leetcodeisgreat"
输出:5
解释:最长子字符串是 "leetc" ,其中包含 2 个 e

示例 3:

输入:s = "bcbcbc"
输出:6
解释:这个示例中,字符串 "bcbcbc" 本身就是最长的,因为所有的元音 a,e,i,o,u 都出现了 0 次。

 

提示:

  • 1 <= s.length <= 5 x 10^5
  • s 只包含小写英文字母。

解法

前缀异或 + 状态压缩。

相似题目:1915. 最美子字符串的数目

Python3

class Solution:
    def findTheLongestSubstring(self, s: str) -> int:
        pos = [inf] * 32
        pos[0] = -1
        vowels = 'aeiou'
        state = ans = 0
        for i, c in enumerate(s):
            for j, v in enumerate(vowels):
                if c == v:
                    state ^= 1 << j
            ans = max(ans, i - pos[state])
            pos[state] = min(pos[state], i)
        return ans

Java

class Solution {

    public int findTheLongestSubstring(String s) {
        int[] pos = new int[32];
        Arrays.fill(pos, Integer.MAX_VALUE);
        pos[0] = -1;
        String vowels = "aeiou";
        int state = 0;
        int ans = 0;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            for (int j = 0; j < 5; ++j) {
                if (c == vowels.charAt(j)) {
                    state ^= (1 << j);
                }
            }
            ans = Math.max(ans, i - pos[state]);
            pos[state] = Math.min(pos[state], i);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findTheLongestSubstring(string s) {
        vector<int> pos(32, INT_MAX);
        pos[0] = -1;
        string vowels = "aeiou";
        int state = 0, ans = 0;
        for (int i = 0; i < s.size(); ++i) {
            for (int j = 0; j < 5; ++j)
                if (s[i] == vowels[j])
                    state ^= (1 << j);
            ans = max(ans, i - pos[state]);
            pos[state] = min(pos[state], i);
        }
        return ans;
    }
};

Go

func findTheLongestSubstring(s string) int {
	pos := make([]int, 32)
	for i := range pos {
		pos[i] = math.MaxInt32
	}
	pos[0] = -1
	vowels := "aeiou"
	state, ans := 0, 0
	for i, c := range s {
		for j, v := range vowels {
			if c == v {
				state ^= (1 << j)
			}
		}
		ans = max(ans, i-pos[state])
		pos[state] = min(pos[state], i)
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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