Leetcode 动态规划 打家劫舍 337 股票 121 122 123 188 子序列 300 674 718 1143 1035
打家劫舍：
337. House Robber III
class Solution {
public:
int rob(TreeNode* root) {
vector res = robTree(root);
return max(res[0], res[1]);
}
vector robTree(TreeNode* cur){
if(cur == NULL) return vector {0, 0};
vector left = robTree(cur->left);
vector right = robTree(cur->right);
//偷了的值
int val1 = cur->val + left[1] + right[1];
int val2 = max(left[1], left[0]) + max(right[1], right[0]);
return {val1, val2};
}
}; 股票问题：
121. Best Time to Buy and Sell Stock
class Solution {
public:
int maxProfit(vector& prices) {
vector> dp(prices.size(), vector(2,0));
dp[0][1] = -prices[0];
for(int i=1; i 122. Best Time to Buy and Sell Stock II
class Solution {
public:
int maxProfit(vector& prices) {
int size = prices.size();
vector> dp(size, vector(2, 0));
dp[0][1] = -prices[0];
for(int i=1; i 123. Best Time to Buy and Sell Stock III
class Solution {
public:
int maxProfit(vector& prices) {
//5种状态
/*
1.无操作
2.第一次持有
3.第一次不持有
4.第二次持有
5.第二次不持有
*/
int size = prices.size();
vector> dp(size, vector(5, 0));
dp[0][1] = dp[0][3] = -prices[0];
for(int i=1; i 188. Best Time to Buy and Sell Stock IV
class Solution {
public:
int maxProfit(int k, vector& prices) {
int size = prices.size();
vector> dp(size, vector(2*k+1, 0));
for(int i=1; i<2*k+1; i++){
if(i%2 == 1)
dp[0][i] = -prices[0];
}
for(int i=1; i 子序列问题：
300. Longest Increasing Subsequence
class Solution {
public:
int lengthOfLIS(vector& nums) {
if(nums.size() == 1) return 1;
vector dp(nums.size(), 1);
int res = 0;
for(int i=1; i nums[j])
dp[i] = max(dp[i], dp[j]+1);
}
if(dp[i] > res){
res = dp[i];
}
}
return res;
}
}; dp[i] 表示下标为i的数字（包括）的前面的最长子序列的长度
因为这里最长子序列不一定在最后，所以要重新设一个值记录最长子序列
初始化的时候所有的值的初始值都应该是1
674. Longest Continuous Increasing Subsequence
class Solution {
public:
int findLengthOfLCIS(vector& nums) {
vector dp(nums.size(), 1);
int res = 1;
for(int i=1; i nums[i-1]){
dp[i] = dp[i-1]+1;
}
if(dp[i] > res) res = dp[i];
}
return res;
}
}; 718. Maximum Length of Repeated Subarray
class Solution {
public:
int findLength(vector& nums1, vector& nums2) {
vector> dp(nums1.size()+1, vector(nums2.size()+1, 0));
//dp数组的含义：nums1中下标为i-1和nums2下标为j-1的数的前面的最长相同子序列
int res = 0;
for(int i=1; i<=nums1.size(); i++){
for(int j=1; j<=nums2.size(); j++){
if(nums1[i-1] == nums2[j-1]){
dp[i][j] = dp[i-1][j-1] +1;
}
if(res < dp[i][j]) res = dp[i][j];
}
}
return res;
}
}; 1143. Longest Common Subsequence
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
vector> dp(text1.size()+1, vector(text2.size()+1, 0));
//在长度text1【0，i-1】和text2长度【0.j-1】中最长的公共子序列
for(int i=1;i<=text1.size(); i++){
for(int j=1; j<=text2.size(); j++){
if(text1[i-1] == text2[j-1]){
dp[i][j] = dp[i-1][j-1]+1;
}else{
dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[text1.size()][text2.size()];
}
}; 一定要注意好下标
1035. Uncrossed Lines
class Solution {
public:
int maxUncrossedLines(vector& nums1, vector& nums2) {
vector> dp(nums1.size()+1, vector(nums2.size()+1, 0));
for(int i=1; i<=nums1.size(); i++){
for(int j=1; j<=nums2.size(); j++){
if(nums1[i-1] == nums2[j-1]){
dp[i][j] = dp[i-1][j-1]+1;
}else{
dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[nums1.size()][nums2.size()];
}
};