LeetCode刷题 | 735. Asteroid Collision12_16
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We are given an array asteroids ofintegers representing asteroids in a row.
For each asteroid, the absolute value represents itssize, and the sign represents its direction (positive meaning right, negativemeaning left). Each asteroid moves at the same speed.
Find out the state of the asteroids after allcollisions. If two asteroids meet, the smaller one will explode. If both arethe same size, both will explode. Two asteroids moving in the same directionwill 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 movingright.
Asteroids moving the same direction never meet, so noasteroids will meet each other.
Note:
· The length of asteroids will beat most 10000.
· Each asteroid will be a non-zero integer in the range [-1000,1000]..
这题花了太久才做出来,很烦,思维不是很清晰,代码结构也不是很简洁,以后在做题的时候记得for循环的时候,当修改i的值的时候最后的那个++要记住
class Solution {
public:
vector<int> asteroidCollision(vector<int>& asteroids) {
vector<int> theNums;
inti,growA,growB;
if(asteroids.size() == 0)
{
theNums.push_back(asteroids[0]);
return theNums;
}
for(i = 0; i < asteroids.size()-1; i++)
{
if(asteroids[i] > 0 && asteroids[i + 1]<0)
{
growA= i; growB = i + 1;
while(growA>=0&&growB<=asteroids.size()-1)
{
if(abs(asteroids[growA]) - abs(asteroids[growB]) < 0)
{
asteroids[growA]= 0; growA--;
while(growA>=0 && asteroids[growA] == 0) growA--;
if(growA < 0 || asteroids[growA] < 0 )
{
i= growB-1;
break;
}
else
{
continue;
}
}
elseif(abs(asteroids[growA]) - abs(asteroids[growB]) > 0)
{
asteroids[growB]= 0;
growB++;
while(growB<asteroids.size()&&asteroids[growB]==0)growB++;
if(growB >= asteroids.size()||asteroids[growB] > 0 )
{
i= growB-1;
break;
}
else
{
continue;
}
}
else
{
asteroids[growB]= 0; asteroids[growA] = 0;
growA--;
growB++;
while(growA >= 0 && asteroids[growA] == 0) growA--;
while(growB<asteroids.size() && asteroids[growB] == 0) growB++;
if(growB >= asteroids.size() || asteroids[growB] > 0)
{
i= growB-1;
break;
}
if(growA < 0 || asteroids[growA] < 0)
{
i= growB-1;
break;
}
continue;
}
}
}
}
for(int i = 0; i < asteroids.size(); i++)
{
if(asteroids[i] != 0)
theNums.push_back(asteroids[i]);
}
return theNums;
}
};
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