385. Mini Parser

Question

385. Mini Parser

Solution

分析：用NI(count,list)来表示NestedInteger，则解析字符串[123,[456,[789]]]过程如下：

# 首先将字符串转化为字符数组，遍历每个字符
[      压栈操作 NI(0, null)
123    给栈顶元素设置值 NI(0, NI(123))
,      不处理
[      压栈操作 NI(0, NI(123)) | NI(0, null)
456    给栈顶元素设置值 NI(0, NI(123)) | NI(0, 456)
,      不处理
[      压栈操作 NI(0, NI(123)) | NI(0, NI(456)) | NI(0, null)
789    给栈顶元素设置值 NI(0, NI(123)) | NI(0, NI(456)) | NI(0, NI(789))
]      出栈并将出栈后的元素添加到栈顶元素NI(0, NI(123)) | NI(0, [NI(456),NI(0, NI(789))])
]      出栈并将出栈后的元素添加到栈顶元素NI(0, [NI(123), NI(0, [NI(456), NI(0, NI(789))])])
]      不做处理
# 栈中最后一个元素就是返回结果

Java实现：

public NestedInteger deserialize(String s) {
    if (!s.startsWith("[")) return new NestedInteger(Integer.parseInt(s));
    Stack<NestedInteger> stack = new Stack<>();
    // 123,[456,[789]]

    char[] arr = s.toCharArray();
    for (int i = 0; i < arr.length; i++) {
        char ch = arr[i];
        if (ch == '[') {
            stack.push(new NestedInteger());
        } else if (ch == ']' & stack.size() > 1) {
            NestedInteger pop = stack.pop();
            stack.peek().add(pop);
        } else if (Character.isDigit(ch) || ch == '-') {
            boolean pos = true;
            if (ch == '-') {
                pos = false;
                i++;
            }
            int num = 0;
            while (i < arr.length && Character.isDigit(arr[i])) {
                num *= 10;
                num += arr[i++] - '0';
            }
            i--;
            stack.peek().add(new NestedInteger(pos ? num : -num));
        }
    }
    return stack.pop();
}

Reference

• https://gitee.com/okokabcd/leetcode