解决两个字符串的动态规划问题，一般都是用两个指针 i,j 分别指向两个字符串的最后，然后一步步往前走，
缩小问题的规模。
1. 最长公共子序列：求两个字符串的最长公共子序列，代码如下：
1 class Solution { 2 public: 3 int longestCommonSubsequence(string text1, string text2) 4 { 5 const int text1_len = text1.size(); 6 const int text2_len = text2.size(); 7 int dp[text1_len+1][text2_len+1]; 8 memset(dp,0,sizeof(dp)); 9 //base case 10 for(int i = 1;i<=text1_len;++i) 11 { 12 dp[i][0] = 0; 13 } 14 for(int j = 1;j<=text2_len;++j) 15 { 16 dp[0][j] = 0; 17 } 18 //state move 19 for(int i = 1;i<=text1_len;++i) 20 { 21 for(int j = 1;j<=text2_len;++j) 22 { 23 if(text1[i-1]==text2[j-1]) 24 { 25 dp[i][j] = dp[i-1][j-1]+1; 26 } 27 else 28 { 29 dp[i][j] = max(dp[i][j-1],dp[i-1][j]); 30 } 31 } 32 } 33 //last state as target 34 return dp[text1_len][text2_len]; 35 } 36 };
2. 编辑距离：求一个字符串 转换到另一个字符串的最少操作次数，代码如下
class Solution { public: int minDistance(string word1, string word2) { if(word1.empty()&&word2.empty()){ return 0; } else if(word1.empty()) { return word2.size(); } else if(word2.empty()) { return word1.size(); } else { const int word1_len = word1.size(); const int word2_len = word2.size(); int dp[word1_len][word2_len];//dp[i][j] word1[0:i]到word2[0:j]的最短编辑距离 // memset(dp,0,sizeof(dp)); for(int i = 0;i<word1.size();++i) { for(int j = 0;j<word2.size();++j) { int one,two,three; //base case if(i==0&&j==0) { one = 0; //dp[i-1][j-1] two = 1;//dp[i-1][j] three = 1;//dp[i][j-1] } else if(i==0) //j!=0 { one = j; //dp[i-1][j-1] two = j+1;//dp[i-1][j] three = dp[i][j-1] ;//dp[i][j-1] } else if(j==0) //j!=0 { one = i; //dp[i-1][j-1] two = dp[i-1][j];//dp[i-1][j] three = i+1;//dp[i][j-1] } // normal case else { one = dp[i-1][j-1]; two = dp[i-1][j]; three = dp[i][j-1]; } //state move if(word1[i]==word2[j]) { dp[i][j] = one; } else { dp[i][j] = min(min(two+1,three+1),one+1); } } } // the last state return dp[word1_len-1][word2_len-1]; } } };