[LeetCode]House Robber II (二次dp)

213. House Robber II

 

 

Total Accepted: 24216 Total Submissions: 80632 Difficulty: Medium

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

这个题目是基于House Robber I 的，所以做这题之前要先知道HouseRobber I的解法。

House Robber I 的传送门:

House Robber I

如果已经ac了第一题，那么这题的意思就是把屋子都改成环状。

在第一题中，dp状态转移方程    max[ i ] = Math.max( max[ i - 1 ], nums[ i - 1 ] + max[ i - 2 ] ) 已经做出来了。那么第二题就很好做了。

第二题中，我的做法就是要再进行一次dp，并且需要记录下选择屋子的首尾，记为start和last.

在记录start的时候，要注意start的状态 : 1.当前面的屋子已经被选择。2.当前面的屋子没有被选择。

所以这里的dp转移方程总结为:

            if( max[ i - 1 ] > nums[ i - 1 ] + max[ i - 2 ] ) {
                max[ i ] = max[ i - 1 ];
                start[ i ] = start[ i - 1 ];
            } else {
                max[ i ] = nums[ i - 1 ] + max[ i - 2 ];
                last = i;
                if( max[ i - 1 ] == max[ i - 2 ] ) {
                    start[ i ] = start[ i - 2 ];
                } else {
                    start[ i ] = start[ i - 2 ] == 0 ? 2 : start[ i - 2 ];
                }
            }

最后判断一下，如果是选择最后一个的时候报警了，进行判断是要选择(1,n)还是(0,n-1)的最大价值

总的思想就是进行两次DP，(1,n)和(0,n-1)分别Dp

Ps:暂时没有想到更好更加简洁的方法。不过我觉得是有的，只是本人愚笨没想到

public class Solution {

    public int rob( int[] nums ) {

        int[] max = new int[ nums.length + 1 ];
        int[] start = new int[ nums.length + 1 ];
        max[ 0 ] = 0;
        start[ 0 ] = 0;
        if( nums == null || nums.length == 0 )
            return 0;
        max[ 1 ] = nums[ 0 ];
        start[ 1 ] = 1;
        int last = 1;
        for( int i = 2; i <= nums.length; i++ ) {
            if( max[ i - 1 ] > nums[ i - 1 ] + max[ i - 2 ] ) {
                max[ i ] = max[ i - 1 ];
                start[ i ] = start[ i - 1 ];
            } else {
                max[ i ] = nums[ i - 1 ] + max[ i - 2 ];
                last = i;
                if( max[ i - 1 ] == max[ i - 2 ] ) {
                    start[ i ] = start[ i - 2 ];
                } else {
                    start[ i ] = start[ i - 2 ] == 0 ? 2 : start[ i - 2 ];
                }
            }
        }
        if( ( last + 1 ) % nums.length == start[ nums.length ] ) {
            int[] tail = new int[nums.length-1];
            System.arraycopy( nums, 1, tail, 0, nums.length-1 );
            int preMax = rob2( tail );
            max[ nums.length ] = Math.max( ( max[ nums.length ] - max[ 1 ] ), max[ nums.length - 1 ] );
            max[ nums.length ] = Math.max( max[ nums.length ], preMax );
        }
        return max[ nums.length ];
    }

    public int rob2( int[] nums ) {

        int[] max = new int[ nums.length + 1 ];
        max[ 0 ] = 0;
        if( nums == null || nums.length == 0 )
            return 0;
        max[ 1 ] = nums[ 0 ];
        for( int i = 2; i <= nums.length; i++ ) {
            max[ i ] = Math.max( max[ i - 1 ], nums[ i - 1 ] + max[ i - 2 ] );
        }
        return max[ nums.length ];
    }

    public static void main( String[] args ) {
        Solution s = new Solution();
        int[] nums = new int[] { 2, 2, 4, 3, 2, 5 };
        System.out.println( s.rob( nums ) );
    }

}