代码中StringBuilder存在什么问题？字符串交替合并功能异常排查

问题背景 #

给定两个字符串word1和word2，需按交替顺序合并（以word1开头），若某字符串更长，将剩余部分追加到合并结果末尾。编写的Java代码如下：

class Solution {
    public String mergeAlternately(String word1, String word2) {
        int length = word1.length() + word2.length();
        StringBuilder ans = new StringBuilder(length);
        int left = 0, right = 0, current = 0;
        System.out.println(ans.length());
        while(current < ans.length()) {
            System.out.println("called");
            if(left >= word1.length()) {
                ans.append(word2.charAt(right));
                right++;
            }
            else if(right >= word2.length()) {
                ans.append(word2.charAt(left));
                left++;
            }
            else if(current % 2 == 0) {
                ans.append(word1.charAt(left));
                left++;
            }
            else {
                ans.append(word2.charAt(right));
                right++;
            }
            current++;
        }
        return ans.toString();
    }
}

测试输入：word1 = "abc", word2 = "pqr"，预期输出为"apbqcr"，但实际输出为空字符串，需排查代码中StringBuilder相关逻辑问题。

核心问题 #

1. StringBuilder循环条件完全错误 #

你用StringBuilder(int length)初始化时，传入的数值是内部缓冲区的初始容量，不是当前字符串的长度。刚创建的StringBuilder是空的，ans.length()返回0，导致current < ans.length()（即0 < 0）不成立，循环根本没有执行，直接返回空字符串。

2. 附带的字符追加错误 #

当right >= word2.length()（即word2已遍历完）时，代码错误地调用ans.append(word2.charAt(left))，这里应该追加word1的剩余字符，即word1.charAt(left)。

修正方案 #

方案一：保留原交替逻辑（按current奇偶控制顺序） #

class Solution {
    public String mergeAlternately(String word1, String word2) {
        int totalLength = word1.length() + word2.length();
        StringBuilder ans = new StringBuilder(totalLength);
        int left = 0, right = 0, current = 0;
        
        // 循环条件改为遍历总长度次数，而非StringBuilder当前长度
        while(current < totalLength) {
            if(left >= word1.length()) {
                ans.append(word2.charAt(right));
                right++;
            }
            else if(right >= word2.length()) {
                // 修正为追加word1的剩余字符
                ans.append(word1.charAt(left));
                left++;
            }
            else if(current % 2 == 0) {
                ans.append(word1.charAt(left));
                left++;
            }
            else {
                ans.append(word2.charAt(right));
                right++;
            }
            current++;
        }
        return ans.toString();
    }
}

方案二：更简洁的交替追加逻辑 #

直接循环直到两个字符串都遍历完毕，依次追加各自的字符：

class Solution {
    public String mergeAlternately(String word1, String word2) {
        int totalLength = word1.length() + word2.length();
        StringBuilder ans = new StringBuilder(totalLength);
        int left = 0, right = 0;
        
        while (left < word1.length() || right < word2.length()) {
            if (left < word1.length()) {
                ans.append(word1.charAt(left++));
            }
            if (right < word2.length()) {
                ans.append(word2.charAt(right++));
            }
        }
        return ans.toString();
    }
}

关键说明 #

• StringBuilder的length()方法返回的是当前已存储的字符数量，初始为空时为0；而构造方法传入的参数只是设置缓冲区初始容量，用于优化性能，不代表当前字符长度。
• 处理剩余字符时，要明确区分哪个字符串还有未遍历的字符，避免追加错误的内容。

内容的提问来源于stack exchange，提问作者21SCS34 Arumugaperumal