标签：dia exce col blog other 否则 require private equals
Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.
Example 1:
Input: "sea", "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Note:
题目含义：给定两个字符串，每次可以删除一个字母，求使得两个字符串相等的最小步骤
方法一：找到最终公共子串，然后计算其余字符长度
private String findMaxCommonSubString(String word1,String word2) { String min = word1.length()<word2.length()?word1:word2; String max = min.equals(word1)?word2:word1; if (max.contains(min)) return min; for (int i=0;i<min.length();i++) { for (int a=0,b=min.length()-i;b!=min.length()+1;a++,b++) { String subStr = min.substring(a,b); if (max.contains(subStr)) return subStr; } } return ""; } public int minDistance(String word1, String word2) { String maxCommonSubStr = findMaxCommonSubString(word1, word2); if (maxCommonSubStr.isEmpty()) return word1.length() + word2.length(); int commonPosi1 = word1.indexOf(maxCommonSubStr); int commonPosi2 = word2.indexOf(maxCommonSubStr); return word1.length() - maxCommonSubStr.length() + word2.length() - maxCommonSubStr.length(); return word1.length() - val + word2.length() - val; }
方法二：问题可以转化为求两个字符串的最长公共子序列（注意不是子字符串）。
首先求最长公共子序列（LCS），然后，用两个字符串的长度分别减去公共子序列的长度，然后再相加即为要删除的长度。
设有二维数组
其中，
此时，
1 public int minDistance(String word1, String word2) { 2 int dp[][] = new int[word1.length()+1][word2.length()+1]; 3 for(int i = 0; i <= word1.length(); i++) { 4 for(int j = 0; j <= word2.length(); j++) { 5 if(i == 0 || j == 0) dp[i][j] = 0; 6 else dp[i][j] = (word1.charAt(i-1) == word2.charAt(j-1)) ? dp[i-1][j-1] + 1 7 : Math.max(dp[i-1][j], dp[i][j-1]); 8 } 9 } 10 int val = dp[word1.length()][word2.length()]; 11 return word1.length() - val + word2.length() - val; 12 }
583. Delete Operation for Two Strings
标签：dia exce col blog other 否则 require private equals
原文地址：http://www.cnblogs.com/wzj4858/p/7686936.html
