难度:Easy
Three stones are on a number line at positions a, b, and c.
Each turn, you pick up a stone at an endpoint (ie., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions x, y, z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an integer position k, with x < k < z and k != y.
The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.
When the game ends, what is the minimum and maximum number of moves that you could have made? Return the answer as an length 2 array: answer = [minimum_moves, maximum_moves]
Example 1:
Input: a = 1, b = 2, c = 5
Output: [1,2]
Explanation: Move the stone from 5 to 3, or move the stone from 5 to 4 to 3.
Example 2:
Input: a = 4, b = 3, c = 2
Output: [0,0]
Explanation: We cannot make any moves.
Example 3:
Input: a = 3, b = 5, c = 1
Output: [1,2]
Explanation: Move the stone from 1 to 4; or move the stone from 1 to 2 to 4.
Note:
1 <= a <= 100
1 <= b <= 100
1 <= c <= 100
a != b, b != c, c != a
需要排序。 最小次数不会超过二，需要判断特殊情况。 最大次数就是所有间隔数之和。
class Solution {
void nsort(int &a, int &b, int &c)
{
if(a<b && b<c) return ;
if(a>b) swap(a,b);
if(a>c) swap(a,c);
if(b>c) swap(b,c);
}
public:
vector<int> numMovesStones(int a, int b, int c) {
nsort(a,b,c);
// cout<<a<<" "<<b<< " "<<c<<endl;
int left=b-a-1;
int right=c-b-1;
vector<int> res(2);
res[1]=left+right;
if(left && right)
res[0]= min(min(left,right),2);
else if(left || right)
res[0]=1;
else
res[0]=0;
// cout<<left<<" "<<right<<endl;
return res;
}
};
执行用时 :0 ms, 在所有 C++ 提交中击败了100.00%的用户
内存消耗 :8.2 MB, 在所有 C++ 提交中击败了100.00%的用户