2022.07.13 _ a day

735. 行星碰撞

题目描述

给定一个整数数组 asteroids,表示在同一行的行星.

对于数组中的每一个元素,其绝对值表示行星的大小,正负表示行星的移动方向（正表示向右移动,负表示向左移动）.每一颗行星以相同的速度移动.

找出碰撞后剩下的所有行星.碰撞规则：两个行星相互碰撞,较小的行星会爆炸.如果两颗行星大小相同,则两颗行星都会爆炸.两颗移动方向相同的行星,永远不会发生碰撞.

 

示例 1：

输入：asteroids = [5,10,-5]
输出：[5,10]
解释：10 和 -5 碰撞后只剩下 10 . 5 和 10 永远不会发生碰撞.

示例 2：

输入：asteroids = [8,-8]
输出：[]
解释：8 和 -8 碰撞后,两者都发生爆炸.

示例 3：

输入：asteroids = [10,2,-5]
输出：[10]
解释：2 和 -5 发生碰撞后剩下 -5 .10 和 -5 发生碰撞后剩下 10 .

 

提示：

• 2 <= asteroids.length <= 104
• -1000 <= asteroids[i] <= 1000
• asteroids[i] != 0

Related Topics

• 栈
• 数组

coding

// 栈模拟
class Solution {
    
    public int[] asteroidCollision(int[] asteroids) {
    
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < asteroids.length; i++) {
    
        	
            if (stack.isEmpty() || stack.peek() * asteroids[i] > 0 || stack.peek() < 0) {
    
	         	// 1.栈为空 
	        	// 2.The planet at the top of the stack moves in the same direction as the next planet
	        	// 3.The planet at the top of the stack moves to the left (Even if the movement is oriented differently,This is equivalent to left facing left,right to right,不会发生碰撞)
                stack.push(asteroids[i]);
            } else {
    
            	// When two stars move towards each other,If the right-facing planet in the stack has a higher velocity,Then the new star explodes directly,Then look at the situation of the next planet
            	// 否则,Planets in the stack must explode
                if (stack.peek() <= - asteroids[i]) {
    
                	// If the speed of the top planet is small,The newly added planets are still considered
                    if (stack.peek() < - asteroids[i]) {
    
                        i --;
                    }
                    // If the speed is the same, the planet at the top of the stack and the planet that needs to be added will die together
                    stack.pop();
                }
            }
        }
        int[] res = new int[stack.size()];
        for (int i = res.length - 1; i >= 0; i--) {
    
            res[i] = stack.pop();
        }
        return res;
    }
}