题目链接： https://leetcode.cn/problems/longest-common-subsequence/
TLE
暴力递归+记忆化版本（基于字符串长度无优化版本）//注意text1[i1-1]==text2[i2-1]
class Solution {
public:int dp[1000][1000];int longestCommonSubsequence(string text1, string text2) {int n=text1.size();int m=text2.size();memset(dp,-1,sizeof(dp));return f(dp,text1,text2,n,m);}int f(int dp[][1000],string t1,string t2,int i1,int i2){//basecaseif(dp[i1][i2]!=-1)return dp[i1][i2];if(i1==0||i2==0){dp[i1][i2]=0;return 0;}try(四种不同可能性）int p1=f(dp,t1,t2,i1-1,i2);int p2=f(dp,t1,t2,i1-1,i2-1);int p3=f(dp,t1,t2,i1,i2-1);int p4= t1[i1-1]==t2[i2-1]? p2+1: p2;int big=max(max(p1,p2),max(p3,p4));dp[i1][i2]=big;return dp[i1][i2];}
};
暴力递归+记忆化（基于字符串长度的优化版本）(感觉上是能通过的，但其实没有）
class Solution {
public:int dp[1005][1005];int longestCommonSubsequence(string text1, string text2) {int n=text1.size();int m=text2.size();memset(dp,-1,sizeof(dp));return f(dp,text1,text2,n,m);}int f(int dp[][1005],string t1,string t2,int i1,int i2){if(dp[i1][i2]!=-1)return dp[i1][i2];if(i1==0||i2==0){dp[i1][i2]=0;return 0;}if(t1[i1-1]==t2[i2-1]){dp[i1][i2]=f(dp,t1,t2,i1-1,i2-1)+1;return dp[i1][i2];}else{dp[i1][i2]=max(f(dp,t1,t2,i1-1,i2),f(dp,t1,t2,i1,i2-1));return dp[i1][i2];}}
};
二维DP
普通版本
class Solution {
public:int dp[1005][1005];int longestCommonSubsequence(string text1, string text2) {int n=text1.size();int m=text2.size();memset(dp,0,sizeof(dp));for(int i=1; i<=n; i++){for(int j=1; j<=m;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[n][m];}};
状压版本
class Solution {
public:int longestCommonSubsequence(string text1, string text2) {string s1;string s2;if(text1.size()>text2.size()){s1=text1;s2=text2;}else{s1=text2;s2=text1;}int n=s1.size();int m=s2.size();vector<int>dp(m+1,0);for(int i=1;i<=n;i++){int left=0;for(int j=1;j<=m;j++){if(s1[i-1]==s2[j-1]){int cur=dp[j];dp[j]=left+1;left=cur;}else{left=dp[j];dp[j]=max(dp[j],dp[j-1]);}}}return dp[m];}};
