
给定两个字符串str1和str2,返回两个字符串的最长公共子序列。
str1 = 1A2B3D4B56 str2 = B1D23CA45B6A
公共子序列123456 ,12C4B6都可以
解法:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
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public int[][] getdp(char[] str1, char[] str2){ int[][] dp = new int[str1.length][str2.length]; dp[0][0] = str[0] == str[1] ? 1 : 0; for (int i = 1; i < str1.length; i++){ dp[i][0] = Math.max(dp[i-1][0], str1[i] == str2[0] ? 1 : 0); } for (int j = 1; i < str2.length; j++){ dp[0][j] = Math.max(dp[0][j-1], str1[0] == str2[j] ? 1 : 0); } for (int i = 1; i < str1.length; i++){ for (int j = 1; j < str2.length; j++){ dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]); if (str1[i] == str2[j]){ dp[i][j] == Math.max(dp[i][j], dp[i-1][j-1]+1); } } } return dp }
public String (String str1, String str2){ if(str1 == null || str2 == null || str1 == "" || str2 == ""){ return ""; } char[] chs1 = str1.toCharArray(); char[] chs2 = str2.toCharArray(); int[][] dp = getdp(chs1, chs2); int m = chs1.length -1; int n = chs2.length -1; char[] res = new char[dp[m][n]]; int index = res.length -1; while(index >= 0){ if (n > 0 && dp[m][n] == dp[m][n-1]){ n--; }else if(m >0 && dp[m][n] == dp[m-1][n]){ m--; }else{ res[index--] = chs1[m] n--; m--; } } }
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