【Leetcode每日一题】判定是否互为字符重排

合集 - 每日一题(3)

1.【Leetcode每日一题】判定是否互为字符重排2023-06-26

2.【每日一题】农场主牛牛2023-06-283.【每日一题】AtCoder Beginner Contest 045 | C - たくさんの数式 题解2023-09-13

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题目

 思路

        思路比较简单，作两个长度为26的数组，用来存储对应序列字母的个数。遍历完成后对比两个数组，根据是否存在相同序列位置字母个数不一样的情况来确定两个字符串是否都是由相同的字母构成的。

        值得一提的是在一开始我忽略了“字母个数必须相同”这一条件。

代码

impl Solution {
    pub fn check_permutation(s1: String, s2: String) -> bool {
        let mut letters_on_s1: [u8; 26] = [0; 26];
        let mut letters_on_s2: [u8; 26] = [0; 26];
        let mut s1 = s1.as_bytes().iter();
        let mut s2 = s2.as_bytes().iter();
 
        let base_: usize = 97;
 
        // 单次遍历获取两个字符串的所有字母并存储其在字母表中序号到 letters_on_s1 和 letters_on_s2 中
        loop {
            match (s1.next(), s2.next()) {
                (Some(point_1), Some(point_2)) => {
                    // println!("{} {}", point_1, point_2);
                    let idx: usize = *point_1 as usize - base_;
                    letters_on_s1[idx] += 1;
                    let idx: usize = *point_2 as usize - base_;
                    letters_on_s2[idx] += 1;
                }, (Some(point_1), None) => {
                    // println!("{}", point_1);
                    let idx: usize = *point_1 as usize - base_;
                    letters_on_s1[idx] += 1;
                }, (None, Some(point_2)) => {
                    // println!("{}", point_2);
                    let idx: usize = *point_2 as usize - base_;
                    letters_on_s2[idx] += 1;
                }, (None, None) => {
                    break;
                }
            }
        }
        
        // 先对两个字母序号表进行异或操作，再对得到的新表逐项作求和操作，确定字母构成是否相同
        for i in 0..26 {
            // println!("{}-{}", letters_on_s1[i], letters_on_s2[i]);
            letters_on_s1[i] = letters_on_s1[i] ^ letters_on_s2[i];
        }
        let mut answer = letters_on_s1[0];
        for i in 1..26 {
            // println!("{}", letters_on_s1[i]);
            answer = answer + letters_on_s1[i];
        }
 
        answer == 0
    }
}