字符串编修

算法想法源自《C语言名题精选百则》。

源码如下：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define M 10

int dealWith(char *s,char *t);
int min3(int a,int b,int c);

int main()
{
	char *s = "xbyzek";
	char *t = "abcdef";
	
	printf("source:");
	puts(s);
	printf("target:");
	puts(t);
	printf("%d\n",dealWith(s,t));
	printf("source:");
	puts(s);
	printf("target:");
	puts(t);

	return 0;
}

int dealWith(char *s,char *t)
{
	int i,j;
	int s_len = strlen(s);
	int t_len = strlen(t);
	int **cost = (int **)malloc(sizeof(int) * max(s_len,t_len) + 1);
	cost[0] = (int *)malloc(sizeof(int) * (s_len + 1)* (t_len + 1));
	for(i = 1;i <= s_len;i++)
		cost[i] = cost[i - 1] + t_len + 1;
	cost[0][0] = 0;
	for(i = 1;i < s_len + 1;i++)
		cost[0][i] = i;
	for(i = 1;i < t_len + 1;i++)
		cost[i][0] = i;
	for(i = 1;i < s_len;i++)
	{
		for(j = 1;j < t_len;j++)
		{
			if(s[i] == t[j])
				cost[i][j] = cost[i - 1][j - 1];
			else
			{
				cost[i][j] = min3(cost[i - 1][j] + 1,cost[i][j - 1] + 1,cost[i - 1][j - 1] + 2);
			}
		}
	}
	return cost[i - 1][j - 1];
}

int min3(int a,int b,int c)
{
	int min = a;

	if(min > b)
		min = b;
	else if(min > c)
		min = c;
	return min;
}