共计 1926 个字符，预计需要花费 5 分钟才能阅读完成。
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
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.
Example 1:
Input: [2,3,2]Output: 3Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.Example 2:
Input: [1,2,3,1]Output: 4Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.
难度：medium
题目：一个专业的盗贼计划抢一条街。每间门店都有一定的现金。这条街的所有店铺以圆圈的方式排列。但是每两个相邻的店铺都有连接的安全系统，如果在一晚个上同时抢了两个相邻的店铺则会触发警报，则抢劫行动结束。
给出一组非负整数代表每间店铺的现金数，盗贼如何在不触发警报的情况下获得最多的钱。
思路：该问题可以转化为 House Robber I，由于数组的头和尾是相连，所以可以转化为 array[0, n – 1] 和 array[1, n] 的问题。然后基于 House Robber I 求最大值。
Runtime: 3 ms, faster than 100.00% of Java online submissions for House Robber II.
class Solution {
public int rob(int[] nums) {
// nums = null
if (null == nums || nums.length < 1) {
return 0;
}
// only 1 element
if (1 == nums.length) {
return nums[0];
}
if (2 == nums.length) {
return Math.max(nums[0], nums[1]);
}
// iterate
int prev = 0, next = nums[0], curVal = next;
for (int i = 1; i < nums.length – 1; i++) {
curVal = Math.max(next, nums[i] + prev);
prev = next;
next = curVal;
}
int result = next;
prev = 0;
next = curVal = nums[1];
for (int i = 2; i < nums.length; i++) {
curVal = Math.max(next, nums[i] + prev);
prev = next;
next = curVal;
}
return Math.max(result, next);
}
}
代码简化
class Solution {
public int rob(int[] nums) {
// nums = null
if (null == nums || nums.length < 1) {
return 0;
}
int n = nums.length;
if (1 == n) {
return nums[0];
}
if (2 == n) {
return Math.max(nums[0], nums[1]);
}
return Math.max(calcMaxValue(nums, 0, n – 1), calcMaxValue(nums, 1, n));
}
private int calcMaxValue(int[] nums, int start, int end) {
int prev = 0, next = nums[start], curVal = next;
for (int i = start + 1; i < end; i++) {
curVal = Math.max(next, nums[i] + prev);
prev = next;
next = curVal;
}
return next;
}
}
