题目地址:https://leetcode.com/problems/minimum-increment-to-make-array-unique/description/

题目描述

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.

Note:

1、 0<=pushed.length==popped.length<=1000
2、 0<=pushed[i],popped[i]<1000
3、 pushedisapermutationofpopped.;
4、 pushedandpoppedhavedistinctvalues.;

题目大意

给出了一个栈的输入数字,按照这个顺序输入到栈里,能不能得到一个对应的栈的输出序列。

解题方法

模拟过程

我使用的方法异常简单粗暴,直接模拟这个过程。

题目已知所有的数字都是不同的。我们在模拟这个弹出的过程中,进行一个判断,如果这个弹出的数字和栈顶数字是吻合的,那么栈就要把已有的数字弹出来。如果栈是空的,或者栈顶数字和弹出的数字不等,那么我们应该把pushed数字一直往里放,直到相等为止。

最后,如果栈的入栈序列能得到这个出栈序列,那么栈应该是空的。

class Solution(object):
    def validateStackSequences(self, pushed, popped):
        """
        :type pushed: List[int]
        :type popped: List[int]
        :rtype: bool
        """
        stack = []
        N = len(pushed)
        pi = 0
        for i in range(N):
            if stack and popped[i] == stack[-1]:
                stack.pop()
            else:
                while pi < N and pushed[pi] != popped[i]:
                    stack.append(pushed[pi])
                    pi += 1
                pi += 1
        return not stack

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

使用C++代码如下,思路和上面一样,其实可以简化代码。

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        bool canbe = true;
        int N = pushed.size();
        stack<int> s;
        int pi = 0;
        for (int i = 0; i < N; i++) {
            s.push(pushed[i]);
            while (!s.empty() && s.top() == popped[pi]) {
                s.pop();
                pi++;
            }
        }
        return s.empty();
    }
};

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

DDKK.COM 弟弟快看-教程,程序员编程资料站,版权归原作者所有

本文经作者:负雪明烛 授权发布,任何组织或个人未经作者授权不得转发