作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
目录
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:
0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed
is a permutation ofpopped
.pushed
andpopped
have distinct values.
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
日期
2018 年 11 月 24 日 —— 周日开始!一周就过去了~