198.打家劫舍
五部曲:
dp数组下标及含义:考虑下标i(包括i)以内的房屋,最多可以偷窃的金额为dp[i]。
dp数组初始化:dp[0] = nums[0];dp[1] = max(nums[0], nums[1]);
递推公式: dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
遍历方向:从前到后遍历
class Solution {
public:int rob(vector<int>& nums) {if (nums.size() == 0)return 0;if (nums.size() == 1)return nums[0];vector<int> dp(nums.size());dp[0] = nums[0];dp[1] = max(nums[0], nums[1]);for (int i = 2; i < nums.size(); i++) {dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);}return dp[nums.size() - 1];}
};
213.打家劫舍II
思路: 本题和上一题的区别就是房屋成环,首尾的房间时相邻的,所以偷首就不能偷尾,偷尾就不能偷首。将题目分成两种情况分别求出结果,取最大结果。
class Solution {
public:int rob(vector<int>& nums) {if (nums.size() == 0)return 0;if (nums.size() == 1)return nums[0];int result1 = robRange(nums, 0, nums.size() - 2);int result2 = robRange(nums, 1, nums.size() - 1);return max(result1, result2);}int robRange(vector<int>& nums, int start, int end) {if (end == start)return nums[start];vector<int> dp(nums.size());dp[start] = nums[start];dp[start + 1] = max(nums[start], nums[start + 1]);for (int i = start + 2; i <= end; i++) {dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);}return dp[end];}
};
337.打家劫舍III
思路: 树形dp,要在二叉树的遍历中进行dp操作。
有些难度,还是需要多看几遍。
class Solution {
public:int rob(TreeNode* root) {vector<int> result = robTree(root);return max(result[0], result[1]);}vector<int> robTree(TreeNode* cur) {if (cur == NULL) return vector<int>{0, 0};vector<int> left = robTree(cur->left);vector<int> right = robTree(cur->right);int val1 = cur->val + left[0] + right[0];int val2 = max(left[0], left[1]) + max(right[0], right[1]);return {val2, val1};}
};