首頁 解題筆記 Leetcode Easy

⭐️

# 題目敘述

Given two strings s and goal , return true if you can swap two letters in s so the result is equal to goal , otherwise, return false .

Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j] .

  • For example, swapping at indices 0 and 2 in "abcd" results in "cbad" .

# Example 1

Input: s = "ab", goal = "ba"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.

# Example 2

Input: s = "ab", goal = "ab"
Output: false
Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.

# Example 3

Input: s = "aa", goal = "aa"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.

# 解題思路

# Solution

import java.util.HashSet;
import java.util.Set;
class Solution {
    public boolean buddyStrings(String s, String goal) {
        int n = s.length();
        
        if (s.equals(goal)) {
            Set<Character> temp = new HashSet<>();
            for (char c : s.toCharArray()) {
                temp.add(c);
            }
            return temp.size() < goal.length(); // Swapping same characters
        } // two string equal
        int i = 0;
        int j = n - 1;
        while (i < j && s.charAt(i) == goal.charAt(i)) {
            i++;
        }
        while (j >= 0 && s.charAt(j) == goal.charAt(j)) {
            j--;
        }
        if (i < j) {
            char[] clone = s.toCharArray();
            char temp = clone[i];
            clone[i] = clone[j];
            clone[j] = temp;
            s = new String(clone);
        } // Find two letter to swap, when two letter on deference string in same index isn't equal
        return s.equals(goal);
    }
}




單字
** **
!! !!
片語 & 搭配詞
!! !!