leetcode#777. Swap Adjacent in LR String

最新推荐文章于 2022-10-02 22:03:37 发布

最新推荐文章于 2022-10-02 22:03:37 发布 · 725 阅读

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本文针对LeetCode上的题目SwapAdjacentinLRString（#777），提出了一种利用栈的数据结构来判断两个由‘L’、‘R’和‘X’组成的字符串是否可以通过一系列规定操作相互转换的方法。该方法通过同步扫描两个字符串并根据字符类型进行入栈或出栈操作，最终以栈是否为空作为判断标准。

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leetcode#777. Swap Adjacent in LR String

Problem Description:

In a string composed of ‘L’, ‘R’, and ‘X’ characters, like “RXXLRXRXL”, a move consists of either replacing one occurrence of “XL” with “LX”, or replacing one occurrence of “RX” with “XR”. Given the starting string start and the ending string end, return True if and only if there exists a sequence of moves to transform one string to the other.

Solution:
This problem is similar to quotes match more or less. The transformation can’t exchange the relative positions between L and R and X is useless. So we can solve the problem by stack easily. We scan the start and end simultaneously. Each time we meet an L in end or a R in start, we push it into the stack. On the other hand, each time we meet an L in start or a R in end we pop it from the stack. The condition that there exists a sequence of moves to transform one string to the other is that if and only if the stack is empty after the scanning.

Code:

class Solution {
public:
    bool canTransform(string start, string end) {
        stack<int> s;
        for (int i=0; i<start.size(); i++) {
            if (end[i] == 'L') s.push(1);
            if (start[i] == 'L' && !s.empty()) {
                if (s.top() == 1) s.pop();
                else return false;
            }
            if (start[i] == 'R') s.push(2);
            if (end[i] == 'R' && !s.empty()) {
                if (s.top() == 2) s.pop();
                else return false;
            }
        }
        return s.empty();
    }
};