Description
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.
Solution
Two-pass iteration, time O(n), space O(1)
把first house和last house分别排除掉计算即可。计算方式和Hose Robber相同。注意处理边界情况。
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
int notRobFirst = rob(nums, 1, nums.length - 1);
int notRobLast = rob(nums, 0, nums.length - 2);
return Math.max(notRobFirst, notRobLast);
}
public int rob(int[] nums, int i, int j) {
int robPrev = 0;
int notRobPrev = 0;
while (i <= j) {
int tmp = robPrev;
robPrev = nums[i] + notRobPrev;
notRobPrev = Math.max(tmp, notRobPrev);
++i;
}
return Math.max(robPrev, notRobPrev);
}
}
