字符串最小变换次数——动态规划

感觉自己算法太菜了。woc，想找一个算法大佬，给他做舔狗。

package offer;

/**
 * 动态规划求解-将字符串A变换为字符串B 所用的最少字符操作次数
 */

public class Day0418_demo01 {

    public static int fun(String s1 , String s2){

        int len1 = s1.length();
        int len2 = s2.length();

        int arr[][] = new int[len1][len2];

        //边界
        for(int i=0; i<len1; i++){
            arr[i][0] = i;
        }
        for(int j=0; j<len2; j++){
            arr[0][j] = j;
        }

        for(int i=1; i<len1; i++){
            for(int j=1; j<len2; j++){
                int a = s1.charAt(i) == s2.charAt(j) ? 0 : 1;
                int d = arr[i-1][j]++;
                int in = arr[i][j-1]++;
                int sub = arr[i-1][j-1] + a;
                arr[i][j] = minNum(d , in , sub);
            }
        }
        System.out.println(arr[len1-1][len2-1]);
        return 0;
    }

    //计算最小值
    public static int minNum(int a, int b, int c) {
        int min = a > b ? b : a;
        return min > c ? c : min;
    }

    public static void main(String args[]){
        fun("asdfwfew" , "jfew");
    }
}

 

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