LeetCode 72. Edit Distance Java

72.Edit Distance(编辑距离)

题目:

  给定两个单词 word1 和 word2,计算出将 word1 转换成 word2 所使用的最少操作数 。

  你可以对一个单词进行如下三种操作:

    1. 插入一个字符
    2. 删除一个字符
    3. 替换一个字符

思路:

  多次选择试图得到最优解,那么考虑动态规划。

  先假设word1有len1位,word2有len2位,建立数组step,step[i][j]就代表我们要将word1前 i 位转换为word2前 j 位的最少数量。

  此时word1查找到第 i+1 位字母a,word2查找到第 j+1 位字母b,我们直接比较这两个字母,能得到两种情况:

  1.a=b:那么就不需要操作,此时word1前 i+1 位替换为word2前 j+1 位只需要step[i][j]步

    即为   step[ i+1 ][ j+1 ] = step[ i ][ j ]

  2.a!=b:就要在给定的三种操作方式中选择最优解,再增加一步操作即可

    即为   step[ i+1 ][ j+1 ] = Min(num1,num2,num3);

  最后得到的step[ i ][ j ]就是最短编辑距离。

图解:

  首先建立数组,将红色部分赋值,之后开始按照顺序计算,从word1转换为word2,step[ i ][ j ]就是最短编辑距离,step[ i-1][ j ]就是插入,step[ i ][ j-1 ]就是删除,step[ i-1 ][ j-1 ]就是修改,如果不相同就从三种决策中寻找最小值加一,相同就直接添加,值等于step[ i-1 ][ j-1 ]。

  例如AB->ABC(大号红色字体),B和C不同,左侧(AB->AB)0次,左上(A->AB)1次,上方(A->ABC)2次,选择三种情况最小值,再进行一步操作(左侧AB->AB->ABC),只需要1次。

  或者ABDC->ABC(大号蓝色字体),C与C相同,直接进行左上(ABD->AB-ABC),也只需要1次。

  

代码:

  

	public static int minDistance(String word1, String word2)
	{
		int len1 = word1.length();
	    int len2 = word2.length();

	    int[][] step = new int[len1 + 1][len2 + 1];

	    for (int i = 0; i <= len1; i++)
	        step[i][0] = i;
	    for (int j = 0; j <= len2; j++)
	        step[0][j] = j;

	    for (int i = 1; i <= len1; i++)
	    {
	        char letter1 = word1.charAt(i-1);
	        for (int j = 1; j <= len2; j++)
	        {
	            char letter2 = word2.charAt(j-1);

	            if (letter1 == letter2)
	            {   //若字母相同,即直接添加,不增加步数
	                step[i][j] = step[i-1][j-1];
	            }
	            else
	            {
	                int ReNum = step[i-1][j-1] + 1;
	                //修改
	                int InNum = step[i-1][j] + 1;
	                //插入
	                int DeNum = step[i][j-1] + 1;
	                //删除
	                int min = Math.min(ReNum,Math.min(InNum, DeNum));
	                step[i][j] = min;
	            }
	        }
	    }
	    return step[len1][len2];
    }

  

    

原文地址:https://www.cnblogs.com/blogxjc/p/10857908.html

时间: 2024-09-28 20:35:20

LeetCode 72. Edit Distance Java的相关文章

leetCode 72.Edit Distance (编辑距离) 解题思路和方法

Edit Distance Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a character b) Delete a char

LeetCode 72 Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a character b) Delete a character c) Repla

第十八周 Leetcode 72. Edit Distance(HARD) O(N^2)DP

Leetcode72 看起来比较棘手的一道题(列DP方程还是要大胆猜想..) DP方程该怎么列呢? dp[i][j]表示字符串a[0....i-1]转化为b[0....j-1]的最少距离 转移方程分三种情况考虑 分别对应三中操作 因为只需要三个值就可以更新dp[i][j] 我们可以把空间复杂度降低到O(n) Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement)); Delet

19.2.13 [LeetCode 72] Edit Distance

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "h

72. Edit Distance (JAVA)

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "h

【leetcode】Edit Distance 详解

下图为TI C6xx DSP Nyquist总线拓扑图,总线连接了master与slave,提供了高速的数据传输.有很多种速率不同的总线,如图中的红色方框,最高速总线为CPU/2 TeraNet SCR(即VBUSM SCR),带宽为256bit,其他低速总线为CPU/3,CPU/6,带宽参考图中所示.总线之间用Bridge(桥)连接,作用包括转换总线的速率,使之与所流向总线的速率相同等. 在具体应用中,各种速率的总线完全可以满足复杂的数据传输,而数据传输的瓶颈往往在于连接总线之间的Bridge

72. Edit Distance &amp;&amp; 161. One Edit Distance

72. Edit Distance Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a characterb) Delete a c

刷题72. Edit Distance

一.题目说明 题目72. Edit Distance,计算将word1转换为word2最少需要的操作.操作包含:插入一个字符,删除一个字符,替换一个字符.本题难度为Hard! 二.我的解答 这个题目一点思路也没,就直接看答案了.用的还是dp算法,dp[n1+1][n2+1]中的dp[i][j]表示将word1的前i位,变为word2的前j位需要的步骤.注意第1行是空,第1列也是空. 1.第一行中,dp[0][i]表示空字符""到word2[0,...,i]需要编辑几次 2.第一列中,d

Baozi Leetcode solution 72. Edit Distance

Problem Statement Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: