Monday, August 29, 2016

Leetcode Elimination Game


http://bookshadow.com/weblog/2016/08/28/leetcode-elimination-game/
There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.
We keep repeating the steps again, alternating left to right and right to left, until a single number remains.
Find the last number that remains starting with a list of length n.
Example:
Input:
n = 9,
1 2 3 4 5 6 7 8 9
2 4 6 8
2 6
6

Output:
6

模拟题,时间复杂度O(log n) 空间复杂度O(1)。
根据题设要求的数字移除过程,可以发现每执行完一趟数字移除操作,列表中剩余相邻数字之间的差都会加倍。
因此解决问题的关键就是找到每一趟数字消除操作之后剩余数字的起点。
def lastRemaining(self, n): """ :type n: int :rtype: int """ a = p = 1 cnt = 0 while n > 1: n /= 2 cnt += 1 p *= 2 if cnt % 2: a += p / 2 + p * (n - 1) else: a -= p / 2 + p * (n - 1) return a

http://zyy1314.com/2016/08/28/leetcode390/
解题思路是每次遍历删除元素时,更新并用head记录数组第一个元素。每次遍历之后,数组相邻元素间隔step都会变成2倍,当数组个数为1时,head就是最后剩下的元素。
那什么时候更新head呢?
  1. 当我们从左边开始遍历删除元素时
  2. 当我们从右边开始遍历元素,并且剩下的数组元素个数为奇数时
例如:
2,4,6,8,10
从10开始移动,会删除10,6,2。数组的第一个元素head会被删除,因此需要更新head
2,4,6,8,10,12
从12从右至左遍历,我们会删除12,8,4,head此时依然是2
public int lastRemaining(int n) {
boolean left = true;
int remain = n;
int step = 1;
int head = 1;
while(remain>1){
if(left||remain%2==1){
head +=step;
}
remain /=2;
left =!left;
step *=2;
}
return head;
}


https://segmentfault.com/a/1190000006739199
    public int lastRemaining(int n) {
        int rest = n, start = 1, step = 2;
        boolean left = true;
        while (rest > 1) {
            rest /= 2;
            if (left) start = start + step * rest - step / 2;
            else start = start - step * rest + step / 2;
            step *= 2;
            left = !left;
        }
        return start;
    }
X. Recursive
https://discuss.leetcode.com/topic/58042/c-1-line-solution-with-explanation
After first elimination, all the rest numbers are even numbers.
Divide by 2, we get a continuous new sequence from 1 to n / 2.
For this sequence we start from right to left as the first elimination.
Then the original result should be two times the mirroring result of lastRemaining(n / 2).
int lastRemaining(int n) {
    return n == 1 ? 1 : 2 * (1 + n / 2 - lastRemaining(n / 2));
}

http://www.cnblogs.com/grandyang/p/5860706.html
发现这道题用递归来做很简单,我们用一个bool型变量left2right,为true表示从左往右,为false表示从右往左遍历。当n为1时,不论从左往右还是从右往左都返回1。如果n大于1,且是从左往右的话,我们返回2倍的对n/2的从右往左的遍历;如果是从右往左的话,稍稍麻烦一些,我们肯定还是要对n/2调用递归函数的,
    int lastRemaining(int n) {
        return help(n, true);    
    }
    int help(int n, bool left2right) {
        if (n == 1) return 1;
        if (left2right) {
            return 2 * help(n / 2, false);
        } else {
            return 1 + n - 2 * (1 + n / 2 - help(n / 2, true));
        }
    }

http://codefang.blogspot.com/2016/08/leetcode-weekly-contest390-elimination.html

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