Rust 解法, 执行用时: 2ms, 内存消耗: 424KB, 提交时间: 2021-07-23

struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    * 
        * @param pushV int整型一维数组 
        * @param popV int整型一维数组 
        * @return bool布尔型
    */
    pub fn IsPopOrder(&self, pushV: Vec<i32>, popV: Vec<i32>) -> bool {
        // write code here
        if pushV.len() != popV.len() {
            return false;
        }
        let mut help:Vec<i32> = vec![];
        let mut pop_v = popV;
        pop_v.reverse();
        for item in pushV {
            help.push(item);
            loop {
                match help.pop().take() {
                    Some(cur) => {
                        if let Some(pop_node) = pop_v.pop().take() {
                            if pop_node != cur {
                                pop_v.push(pop_node);
                                help.push(cur);
                                break;
                            }
                        } else {
                            return false;
                        }
                    },
                    None => {
                        break;
                    }
                }
            }
        }
        help.is_empty()


    }
}

C++ 解法, 执行用时: 2ms, 内存消耗: 556KB, 提交时间: 2022-02-10

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        int size = pushV.size();
        stack<int> stk;
        int j = 0;
        for(int i = 0; i < size; ++i){
            while(j < size && (stk.empty() || stk.top() != popV[i])){
                stk.push(pushV[j]);
                ++j;
            }
            if(stk.top() == popV[i]){
                stk.pop();
            } else {
                return false;
            }
        }
        return true;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 556KB, 提交时间: 2021-10-31

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        stack<int> st;
        int pushi = 0;
        int popi = 0;
        
        
        while (pushi < pushV.size())
        {
            st.push(pushV[pushi]);
            ++pushi;
            
            while (!st.empty() && st.top() == popV[popi])
            {
                st.pop();
                ++popi;
            }
        }
        
        return st.empty();
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 560KB, 提交时间: 2021-11-18

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        stack<int> st;
        int i = 0, j = 0;
        while(i < pushV.size()){
            if(pushV[i] == popV[j]){
                ++i;
                ++j;
                while(!st.empty() && popV[j]==st.top()){
                    st.pop();
                    ++j;
                }
            }else{
                st.push(pushV[i++]);
            }
        }
        return st.empty();
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 592KB, 提交时间: 2022-02-10

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        stack<int>stack1;
        int j = 0;
        for(int i = 0;i < pushV.size();i++) {
            stack1.push(pushV[i]);
            while(!stack1.empty() && popV[j] == stack1.top()) {
                stack1.pop();
                j++;
            }
        }
        
        if(stack1.empty())
            return true;
        else
            return false;
    }
};