简单题的五种解法 : 删除字符串中相邻重复项 | Java 刷题打卡

输入："abbaca"
输出："ca"
解释：
例如，在 "abbaca" 中，我们可以删除 "bb" 由于两字母相邻且相同，这是此时唯一可以执行删除操作的重复项。之后我们得到字符串 "aaca"，其中又只有 "aa" 可以执行重复项删除操作，所以最后的字符串为 "ca"。
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class Solution {
    public String removeDuplicates(String s) {
        char[] cs = s.toCharArray();
        Deque<Character> d = new ArrayDeque<>();
        for (char c : cs) {
            if (!d.isEmpty() && d.peekLast().equals(c)) {
                d.pollLast();
            } else {
                d.addLast(c);
            }
        }
        StringBuilder sb = new StringBuilder();
        while (!d.isEmpty()) sb.append(d.pollLast());
        sb.reverse();
        return sb.toString();
    }
}
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class Solution {
    public String removeDuplicates(String s) {
        char[] cs = s.toCharArray();
        char[] d = new char[s.length()];
        int hh = 0, tt = -1;
        for (char c : cs) {
            if (hh <= tt && d[tt] == c) {
                tt--;
            } else {
                d[++tt] = c;
            }
        }  
        StringBuilder sb = new StringBuilder();
        while (hh <= tt) sb.append(d[tt--]);
        sb.reverse();
        return sb.toString();
    }
}
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class Solution {
    public String removeDuplicates(String s) {
        char[] cs = s.toCharArray();
        Deque<Character> d = new ArrayDeque<>();
        for (char c : cs) {
            if (!d.isEmpty() && d.peekLast().equals(c)) {
                d.pollLast();
            } else {
                d.addLast(c);
            }
        }
        StringBuilder sb = new StringBuilder();
        while (!d.isEmpty()) sb.append(d.pollFirst());
        return sb.toString();
    }
}
复制代码

class Solution {
    public String removeDuplicates(String s) {
        char[] cs = s.toCharArray();
        char[] d = new char[s.length()];
        int hh = 0, tt = -1;
        for (char c : cs) {
            if (hh <= tt && d[tt] == c) {
                tt--;
            } else {
                d[++tt] = c;
            }
        }  
        StringBuilder sb = new StringBuilder();
        while (hh <= tt) sb.append(d[hh++]);
        return sb.toString();
    }
}
复制代码

class Solution {
    public String removeDuplicates(String s) {
        char[] cs = s.toCharArray();
        char[] d = new char[s.length()];
        int hh = 0, tt = -1;
        for (char c : cs) {
            if (hh <= tt && d[tt] == c) {
                tt--;
            } else {
                d[++tt] = c;
            }
        }  
        return new String(d, 0, tt + 1);
    }
} 
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