LeetCode解题报告（301)-- 946. Validate Stack Sequences

Problem

Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.

Example 1:

Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
Output: true
Explanation: We might do the following sequence:
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1

Example 2:

Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
Output: false
Explanation: 1 cannot be popped before 2.

Constraints:

• 0 <= pushed.length == popped.length <= 1000
• 0 <= pushed[i], popped[i] < 1000
• pushed is a permutation of popped.
• pushed and popped have distinct values.

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Analysis

  这是一道和stack相关的题目。题目给出两个数组pushed和popped，要求我们校验能否从pushed数组，通过一个stack，产生出popped数组。思路也很清晰，直接用一个stack模拟即可。把pushed中的元素逐个放入到stack中，每次放入后，和popped中的做对比，如果栈顶元素和popped中的相同，元素出栈，popped的下标自增，如果到最后栈为空并且popped的下标到了末尾，说明所有的元素都能处理完，return true；否则return false。

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Solution

  无

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Code

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        stack<int> st;
        int size = pushed.size();
        int idx1 = 0, idx2 = 0;
        while (idx1 != size) {
            st.push(pushed[idx1++]);
            while (!st.empty() && st.top() == popped[idx2]) {
                st.pop();
                idx2++;
            }
        }
        return st.empty() && idx2 == size;
    }
};

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Summary

  这道题目是栈的利用，它从另一个角度考察了我们对栈的理解。这道题目的分享到这里，谢谢您的支持！