第九章 动态规划part07
今天就是打家劫舍的一天,这个系列不算难,大家可以一口气拿下。
198.打家劫舍
视频讲解:https://www.bilibili.com/video/BV1Te411N7SX
// 动态规划
class Solution {public int rob(int[] nums) {if (nums == null || nums.length == 0) return 0;if (nums.length == 1) return nums[0];int[] dp = new int[nums.length];dp[0] = nums[0];dp[1] = Math.max(dp[0], nums[1]);for (int i = 2; i < nums.length; i++) {dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);}return dp[nums.length - 1];}
}// 使用滚动数组思想,优化空间
// 分析本题可以发现,所求结果仅依赖于前两种状态,此时可以使用滚动数组思想将空间复杂度降低为3个空间
class Solution {public int rob(int[] nums) {int len = nums.length;if (len == 0) return 0;else if (len == 1) return nums[0];else if (len == 2) return Math.max(nums[0],nums[1]);int[] result = new int[3]; //存放选择的结果result[0] = nums[0];result[1] = Math.max(nums[0],nums[1]);for(int i=2;i<len;i++){result[2] = Math.max(result[0]+nums[i],result[1]);result[0] = result[1];result[1] = result[2];}return result[2];}
}// 进一步对滚动数组的空间优化 dp数组只存与计算相关的两次数据
class Solution {public int rob(int[] nums) {if (nums.length == 1) {return nums[0];}// 初始化dp数组// 优化空间 dp数组只用2格空间 只记录与当前计算相关的前两个结果int[] dp = new int[2];dp[0] = nums[0];dp[1] = Math.max(nums[0],nums[1]);int res = 0;// 遍历for (int i = 2; i < nums.length; i++) {res = Math.max((dp[0] + nums[i]) , dp[1] );dp[0] = dp[1];dp[1] = res;}// 输出结果return dp[1];}
}
213.打家劫舍II
视频讲解:https://www.bilibili.com/video/BV1oM411B7xq
https://programmercarl.com/0213.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DII.html
class Solution {public int rob(int[] nums) {if (nums == null || nums.length == 0)return 0;int len = nums.length;if (len == 1)return nums[0];return Math.max(robAction(nums, 0, len - 1), robAction(nums, 1, len));}int robAction(int[] nums, int start, int end) {int x = 0, y = 0, z = 0;for (int i = start; i < end; i++) {y = z;z = Math.max(y, x + nums[i]);x = y;}return z;}
}
337.打家劫舍III
视频讲解:https://www.bilibili.com/video/BV1H24y1Q7sY
class Solution {// 1.递归去偷,超时public int rob(TreeNode root) {if (root == null)return 0;int money = root.val;if (root.left != null) {money += rob(root.left.left) + rob(root.left.right);}if (root.right != null) {money += rob(root.right.left) + rob(root.right.right);}return Math.max(money, rob(root.left) + rob(root.right));}// 2.递归去偷,记录状态// 执行用时:3 ms , 在所有 Java 提交中击败了 56.24% 的用户public int rob1(TreeNode root) {Map<TreeNode, Integer> memo = new HashMap<>();return robAction(root, memo);}int robAction(TreeNode root, Map<TreeNode, Integer> memo) {if (root == null)return 0;if (memo.containsKey(root))return memo.get(root);int money = root.val;if (root.left != null) {money += robAction(root.left.left, memo) + robAction(root.left.right, memo);}if (root.right != null) {money += robAction(root.right.left, memo) + robAction(root.right.right, memo);}int res = Math.max(money, robAction(root.left, memo) + robAction(root.right, memo));memo.put(root, res);return res;}// 3.状态标记递归// 执行用时:0 ms , 在所有 Java 提交中击败了 100% 的用户// 不偷:Max(左孩子不偷,左孩子偷) + Max(右孩子不偷,右孩子偷)// root[0] = Math.max(rob(root.left)[0], rob(root.left)[1]) +// Math.max(rob(root.right)[0], rob(root.right)[1])// 偷:左孩子不偷+ 右孩子不偷 + 当前节点偷// root[1] = rob(root.left)[0] + rob(root.right)[0] + root.val;public int rob3(TreeNode root) {int[] res = robAction1(root);return Math.max(res[0], res[1]);}int[] robAction1(TreeNode root) {int res[] = new int[2];if (root == null)return res;int[] left = robAction1(root.left);int[] right = robAction1(root.right);res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);res[1] = root.val + left[0] + right[0];return res;}
}