leetcode 735. Asteroid Collision 行星碰撞 + 类似括号配对 + 栈的典型应用

妖狐艹你老母 2022-06-03 09:44 175阅读 0赞

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 \[-1000, 1000\]..

本题有点类似括号配对，直接使用栈即可完成，

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution 
    {
    public:
        vector<int> asteroidCollision(vector<int>& a)
        {
            vector<int> res;
            int i = 0;
            while (i < a.size())
            {
                if (res.empty() || a[i]>0 || res.back()<0)
                    res.push_back(a[i++]);
                else
                {
                    if (res.back() < -a[i])
                        res.pop_back();
                    else if (res.back() > -a[i])
                        i++;
                    else
                    {
                        res.pop_back();
                        i++;
                    }
                }
            }
            return res;
        }
    };