LeetCode每日一题(1754. Largest Merge Of Two Strings)

You are given two strings word1 and word2. You want to construct a string merge in the following way: while either word1 or word2 are non-empty, choose one of the following options:

If word1 is non-empty, append the first character in word1 to merge and delete it from word1.
For example, if word1 = “abc” and merge = “dv”, then after choosing this operation, word1 = “bc” and merge = “dva”.
If word2 is non-empty, append the first character in word2 to merge and delete it from word2.
For example, if word2 = “abc” and merge = “”, then after choosing this operation, word2 = “bc” and merge = “a”.
Return the lexicographically largest merge you can construct.

A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character strictly larger than the corresponding character in b. For example, “abcd” is lexicographically larger than “abcc” because the first position they differ is at the fourth character, and d is greater than c.

Example 1:

Input: word1 = “cabaa”, word2 = “bcaaa”
Output: “cbcabaaaaa”

Explanation: One way to get the lexicographically largest merge is:

• Take from word1: merge = “c”, word1 = “abaa”, word2 = “bcaaa”
• Take from word2: merge = “cb”, word1 = “abaa”, word2 = “caaa”
• Take from word2: merge = “cbc”, word1 = “abaa”, word2 = “aaa”
• Take from word1: merge = “cbca”, word1 = “baa”, word2 = “aaa”
• Take from word1: merge = “cbcab”, word1 = “aa”, word2 = “aaa”
• Append the remaining 5 a’s from word1 and word2 at the end of merge.

Example 2:

Input: word1 = “abcabc”, word2 = “abdcaba”
Output: “abdcabcabcaba”

Constraints:

• 1 <= word1.length, word2.length <= 3000
• word1 and word2 consist only of lowercase English letters.

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与前两天做的那个 hard 的题目类似， 只不过那个题目拆解成了几个小问题， 这题与其中的一个小问题是一样的。逐个字符的比较两个字符串， 大于或者小于都好处理， 唯独等于的比较复杂一点， 需要看后面的字符， 如果比较到最后都一样则比较两个字符串的长度， 我们认为长的那个更大， 因为我们从长的这个拿字符，下一次比较时可以引入新的字符进行比较

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impl Solution {
    fn is_greater(word1: &[char], word2: &[char]) -> bool {
        let mut i = 0;
        while i < word1.len() && i < word2.len() {
            if word1[i] > word2[i] {
                return true;
            }
            if word1[i] < word2[i] {
                return false;
            }
            i += 1;
        }
        if word1.len() > word2.len() {
            return true;
        }
        false
    }

    pub fn largest_merge(word1: String, word2: String) -> String {
        let chars1: Vec<char> = word1.chars().collect();
        let chars2: Vec<char> = word2.chars().collect();
        let mut i = 0;
        let mut j = 0;
        let mut merge = Vec::new();
        while i < chars1.len() || j < chars2.len() {
            if Solution::is_greater(&chars1[i..], &chars2[j..]) {
                merge.push(chars1[i]);
                i += 1;
                continue;
            }
            merge.push(chars2[j]);
            j += 1;
        }
        merge.into_iter().collect()
    }
}

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