Increasing/ Decreasing Stack

对于此类问题：

对于元素nums[i]，找出往左／右走第一个比它小／大的数字

我们常常用递增栈／递减栈实现。

递增栈实现第一个比它小

递减栈实现第一个比它大

Example: 2　　1　　5　　6　　2　　3

stack: 保证栈是递增的顺序，因此每个数进来之前先删除栈里比它大的数字。

　　　 因此每个数，当它被pop出来时，它在栈里的前一个元素是左边第一个比它小的数，将它pop出来的数是右边第一个比它小的数。

(1) 2

(2) 1　　(2被1pop出来，2左边第一个比它小的没有，右边第一个比它小的是1)

(3) 1　　5

(4) 1　　5　　6

(5) 1　　2　　　　(5, 6 被 2 pop出来。对于5，左边第一个比它小的是1，右边第一个比它小的是2。对于6，左边第一个比它小的是5，右边第一个比它小的是2)

对于每个元素，因为只进栈出栈一次，所以总体的时间复杂度是O(n)

时间： 2024-12-28 21:10:58

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