Stack
- 特性：LIFO （Last In First Out）
- 适用于需要记录之前的状态，必要的时候可以回到之前的状态，或者利用之前的值
- 不像array，不能用index访问，只能每次拿栈顶元素
题外话：动态规划 Dynamic Programming
DP: 记录之前所有状态，随时可以访问任何一个子问题，所以通常用Array或者HashTable，而且不会回到之前的状态，只会利用之前的值
Stack：每次只需要栈顶元素，并且每个状态只会被用O(1)次
Stack Principle
定义好Stack的含义
- 在arr[i]左侧所有比arr[i]大的数
- 递归之前的函数状态（Call Stack）
例题
- Daily Temperature
Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.
Example 1:
ini
Input: temperatures = [73,74,75,71,69,72,76,73]
Output: [1,1,4,2,1,1,0,0]Example 2:
ini
Input: temperatures = [30,40,50,60]
Output: [1,1,1,0]Example 3:
ini
Input: temperatures = [30,60,90]
Output: [1,1,0]Constraints:
1 <= temperatures.length <= 10530 <= temperatures[i] <= 100
Solution:
Find the distance to the next greater element for each arr[i]
Stack Definition: All elements(index) to the right of arr[i] that are greater than arr[i] (monotone increasing stack)
Top of Stack: Next Greater element to the right of arr[i]
High Level Idea:
-
Initialize the stack
-
For each element arr[i] backwards (pop until stack is empty or top of stack > arr[i])
-
Calculate the distance from arr[i] to top of the stack
-
Repeat
class Solution { public int[] dailyTemperatures(int[] temperatures) { int n = temperatures.length; int[] res = new int[n]; Deque stack = new ArrayDeque<>(n); for(int i = n - 1; i >=0; i--) { while(!stack.isEmpty() && temperatures[i] >= temperatures[stack.peek()]) { stack.pop(); } if(stack.isEmpty()) { res[i] = 0; } else { res[i] = stack.peek() - i; } stack.push(i); } return res; }}
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Asteroid Collision
We are given an array asteroids of integers representing asteroids in a row.
For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.
Example 1:
ini
Input: asteroids = [5,10,-5]
Output: [5,10]
Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.Example 2:
vbnet
Input: asteroids = [8,-8]
Output: []
Explanation: The 8 and -8 collide exploding each other.Example 3:
ini
Input: asteroids = [10,2,-5]
Output: [10]
Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.Constraints:
2 <= asteroids.length <= 104-1000 <= asteroids[i] <= 1000asteroids[i] != 0
Solution:
Stack Definition: All asteroid left so far
Top of Stack: Closest survived asteroid to the left of arr[i]
High Level Idea:
- Initialize Stack
- For each arr[i]:
a. if arr[i] > 0, push to stack
b. else keep popping "smaller" until stack is empty or top element < 0
c. special dealing with "equal"
d. push arr[i] to stack if survived
scss
class Solution { public int[] asteroidCollision(int[] asteroids) { Deque<Integer> stack = new ArrayDeque<>(); for(int ast : asteroids) { if(ast > 0) { stack.push(ast); } else { while(!stack.isEmpty() && stack.peek() > 0 && stack.peek() < - ast) { stack.pop(); } if(!stack.isEmpty() && stack.peek() == -ast) { stack.pop(); } else if(stack.isEmpty() || stack.peek() < 0) { stack.push(ast); } } } int[] res = new int[stack.size()]; for(int i = res.length - 1; i >=0; i--) { res[i] = stack.pop(); } return res; }}时间复杂度：O(n)