leetcode 72. 编辑距离

解决思路：

动态规划

状态数组的含义 dp[i][j]: word1的前i个字符--->word2的前j个字符的最小步数

dp[i][j]=  dp[i-1][j-1]                                         word1[i]==word2[j]
            min(dp[i][j-1],dp[i-1][j],dp[i-1][j-1])+1            word1[i]!=word2[j]

代码：

public static void main(String[] args) {
        String word1 = "horse";
        String word2 = "ros";
        int num = minDistance(word1, word2);
        System.out.println(num);
    }

    /**
     * @Description: 状态数组的含义 dp[i][j]: word1的前i个字符--->word2的前j个字符的最小步数
     *  dp[i][j]=  dp[i-1][j-1]                                         word1[i]==word2[j]
     *             min(dp[i][j-1],dp[i-1][j],dp[i-1][j-1])+1            word1[i]!=word2[j]
     * @Date: 2020/2/6 10:39
     * @Author: fuguowen
     * @Return 
     * @Throws 
     */
    public static int minDistance(String word1, String word2) {
        int i=word1.length();
        int j=word2.length();
        if(i==0){
            return j;
        }
        if(j==0){
            return i;
        }
        //先处理左边界和上边界，然后递推
        int[][] arr=new int[i+1][j+1];
        arr[0][0]=0;
        //不断添加单词
        for(int p=1;p<=j;p++){
            arr[0][p]=arr[0][p-1]+1;
        }
        //不断删除单词
        for(int q=1;q<=i;q++){
            arr[q][0]=arr[q-1][0]+1;
        }

        for(int p=1;p<=i;p++){
            for(int q=1;q<=j;q++){
                if(word1.charAt(p-1)==word2.charAt(q-1)){
                    arr[p][q]=arr[p-1][q-1];
                }else{
                    arr[p][q]=Math.min(Math.min(arr[p-1][q-1],arr[p][q-1]),arr[p-1][q])+1;
                }
            }
        }

        return arr[i][j];
    }