力扣解法汇总1662. 检查两个字符串数组是否相等

目录链接:

力扣编程题-解法汇总_分享+记录-CSDN博客

GitHub同步刷题项目:

https://github.com/September26/java-algorithms

原题链接:力扣


描述:

给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同,返回 true ;否则,返回 false 。

数组表示的字符串 是由数组中的所有元素 按顺序 连接形成的字符串。

示例 1:

输入:word1 = ["ab", "c"], word2 = ["a", "bc"]
输出:true
解释:
word1 表示的字符串为 "ab" + "c" -> "abc"
word2 表示的字符串为 "a" + "bc" -> "abc"
两个字符串相同,返回 true

示例 2:

输入:word1 = ["a", "cb"], word2 = ["ab", "c"]
输出:false

示例 3:

输入:word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
输出:true

提示:

  • 1 <= word1.length, word2.length <= 103
  • 1 <= word1[i].length, word2[i].length <= 103
  • 1 <= sum(word1[i].length), sum(word2[i].length) <= 103
  • word1[i] 和 word2[i] 由小写字母组成

解题思路:

* 解题思路:
* 直接拼成字符串比较就好了
 
  

代码:

public class Solution1662 {

    public boolean arrayStringsAreEqual(String[] word1, String[] word2) {
        StringBuilder builder = new StringBuilder();
        String str1;
        for (String str : word1) {
            builder.append(str);
        }
        str1 = builder.toString();
        builder.setLength(0);
        for (String str : word2) {
            builder.append(str);
        }
        return str1.equals(builder.toString());
    }
}

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