213:打家劫舍 II
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LeetCode/src/main/java/leetcode/editor/cn/HouseRobberIi.java
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LeetCode/src/main/java/leetcode/editor/cn/HouseRobberIi.java
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//你是一个专业的小偷,计划偷窃沿街的房屋,每间房内都藏有一定的现金。这个地方所有的房屋都 围成一圈 ,这意味着第一个房屋和最后一个房屋是紧挨着的。同时,相邻的
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//房屋装有相互连通的防盗系统,如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警 。
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//
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// 给定一个代表每个房屋存放金额的非负整数数组,计算你 在不触动警报装置的情况下 ,能够偷窃到的最高金额。
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//
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//
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//
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// 示例 1:
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//
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//
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//输入:nums = [2,3,2]
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//输出:3
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//解释:你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的。
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//
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//
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// 示例 2:
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//
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//
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//输入:nums = [1,2,3,1]
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//输出:4
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//解释:你可以先偷窃 1 号房屋(金额 = 1),然后偷窃 3 号房屋(金额 = 3)。
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// 偷窃到的最高金额 = 1 + 3 = 4 。
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//
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// 示例 3:
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//
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//
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//输入:nums = [0]
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//输出:0
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//
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//
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//
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//
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// 提示:
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//
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//
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// 1 <= nums.length <= 100
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// 0 <= nums[i] <= 1000
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//
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// Related Topics 动态规划
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// 👍 531 👎 0
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package leetcode.editor.cn;
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import java.util.Arrays;
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//213:打家劫舍 II
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public class HouseRobberIi {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new HouseRobberIi().new Solution();
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System.out.println(solution.rob(new int[]{1, 2, 3, 1}));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int rob(int[] nums) {
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int length = nums.length;
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if (length == 1) {
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return nums[0];
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}
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if (length == 2) {
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return Math.max(nums[0], nums[1]);
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}
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return Math.max(range(Arrays.copyOfRange(nums, 0, length - 1)), range(Arrays.copyOfRange(nums, 1, length)));
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}
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public int range(int[] nums) {
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int length = nums.length;
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int start = nums[0],end = Math.max(nums[0],nums[1]);
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for (int i = 2; i < length; i++) {
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int temp = end;
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end = Math.max(start+nums[i],end);
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start = temp;
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}
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return end;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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LeetCode/src/main/java/leetcode/editor/cn/HouseRobberIi.md
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LeetCode/src/main/java/leetcode/editor/cn/HouseRobberIi.md
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<p>你是一个专业的小偷,计划偷窃沿街的房屋,每间房内都藏有一定的现金。这个地方所有的房屋都 <strong>围成一圈</strong> ,这意味着第一个房屋和最后一个房屋是紧挨着的。同时,相邻的房屋装有相互连通的防盗系统,<strong>如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警</strong> 。</p>
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<p>给定一个代表每个房屋存放金额的非负整数数组,计算你 <strong>在不触动警报装置的情况下</strong> ,能够偷窃到的最高金额。</p>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [2,3,2]
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<strong>输出:</strong>3
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<strong>解释:</strong>你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [1,2,3,1]
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<strong>输出:</strong>4
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<strong>解释:</strong>你可以先偷窃 1 号房屋(金额 = 1),然后偷窃 3 号房屋(金额 = 3)。
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偷窃到的最高金额 = 1 + 3 = 4 。</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [0]
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<strong>输出:</strong>0
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 <= nums.length <= 100</code></li>
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<li><code>0 <= nums[i] <= 1000</code></li>
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</ul>
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<div><div>Related Topics</div><div><li>动态规划</li></div></div>\n<div><li>👍 531</li><li>👎 0</li></div>
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