LeetCode 735 Asteroid Collision (栈)

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:

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:

Input: asteroids = [8, -8]Output: []Explanation: The 8 and -8 collide exploding each other.

Example 3:

Input: asteroids = [10, 2, -5]Output: [10]Explanation: The 2 and -5 collide resulting in -5.  The 10 and -5 collide resulting in 10.

Example 4:

Input: asteroids = [-2, -1, 1, 2]Output: [-2, -1, 1, 2]Explanation: The -2 and -1 are moving left, while the 1 and 2 are moving right.Asteroids moving the same direction never meet, so no asteroids will meet each other.

Note:

The length of asteroids will be at most 10000.

Each asteroid will be a non-zero integer in the range [-1000, 1000]..

题目链接：https://leetcode.com/problems/asteroid-collision/description/

题目分析：当前状态需要根据与之前的一些数字比较得到，因此容易想到用栈来维护状态，几种情况下可以直接进栈

1）栈为空； 2）栈顶为负数（此时栈内元素已不会影响到后面）；3）栈顶元素和当前数字同号

需要特殊处理的是栈顶为正数，当前数字为负数，假设分别为s[top]和num，s[top]>=-num && s[top] > 0时栈顶出栈，相等的情况需要单独处理

击败了98.34%

class Solution {        public static boolean isSameSign(int a, int b) {        return a * b > 0;    }        public int[] asteroidCollision(int[] asteroids) {        int n = asteroids.length, top = 0;        int[] stk = new int[n + 1];        for (int i = 0; i < n; i++) {            if (top == 0 || stk[top] < 0 || isSameSign(asteroids[i], stk[top])) {                stk[++ top] = asteroids[i];            } else {                while (top > 0 && stk[top] > 0 && stk[top] < -asteroids[i]) {                    top --;                }                if (top == 0 || stk[top] < 0) {                    stk[++ top] = asteroids[i];                } else {                    if (stk[top] == -asteroids[i]) {                        top --;                    }                }            }        }        int[] ans = new int[top];        for (int i = 1; i <= top; i++) {            ans[i - 1] = stk[i];        }        return ans;    }}