kmp模板匹配next数组求解

求字符串的next的数组，只需要掌握三个要点就可以：

（1）初始化：k=0，j=1，next[0] = -1,next[1] =0;//k表示前串起始位置，j表示后串起始位置

（2）前串和后串比较相等的时候，j,k,next分别怎么变化；

（3）前串和后串比较不相等的时候，j,k,next分别怎么变化；

*****1，初始化就不多说了，按照规则初始化就可以了;

*****2，当前串和后串比较相等的时候，即s[k] == s[j]时，所做的操作是：j++,k++,然后是next[j] = k,然后继续比较；

*****3，当前串和后串比较不相等的时候，即s[k] != s[j]时，需要分情况而定：

（1）当k = 0的时候，j++,然后next[j] = k;

（2）当k ！= 0的时候，k = next[k]，接着比较s[k]与s[j]的大小关系，重复操作不相等的操作，直到next更新为止；

下面进行实例操作，看不懂的可以对例子进行研究：

针对字符串s = "aabccaaabaac"

首先初始化：

next[0]=-1; next[1]=0;

k=0; j=1;

k = 0,j = 1,s[k] = s[0] = 'a',s[j] = s[1] = 'a',即s[k] == s[j],所以 j++,k++,即 j = 2,k = 1,next[j] = net[2] = k = 1，即next[2] = 1;

next[0]=-1; next[1]=0; next[2]=1;

k=1; j=2;

k = 1,j = 2, s[k] = s[1] = 'a',s[j] = s[2] = 'b' ,即s[k] != s[j]，且k != 0,所以k = nxet[k] = next[1] = 0,k = 0,接着比较s[k]与s[j]的大小,即

k =0,j = 2,s[k] = s[0] = 'a',s[j] = s[2] = 'b'，即s[k] != s[j]，且k = 0，所以j++，即j = 3,next[j] = next[3] = k = 0，即next[3] = 0;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0;

k=0; j=3;

k = 0,j = 3,s[k] = s[0] = 'a',s[j] = s[3] = 'c'，即s[k] != s[j]，且k =0，所以j++，即j = 4,next[j] = next[4] = k = 0，即next[4] = 0; 

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0;

k=0; j=4;

k = 0,j = 4,s[k] = s[0] = 'a',s[j] = s[4] = 'a'，即s[k] = s[j]，所以j++,k++，即k = 1,j = 5,next[j] = next[5] = k = 1，即next[5] =1;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1;

k=1; j=5;

k = 1,j = 5,s[k] = s[1] = 'a',s[j] = s[5] = 'a'，即s[k] = s[j]，所以j++,k++，即k = 2,j = 6,next[j] = next[6] = k = 2，即next[6] =2;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1; next[6] =2;

k=2; j=6;

k = 2,j = 6,s[k] = s[2] = 'b',s[j] = s[6] = 'a'，即s[k] != s[j]，且k != 0，所以k = next[k] = next[2] = 1，即k = 1,接着比较s[k]与s[j]，即

k =1,j = 6,s[k]  = s[1] = 'a',s[j] = s[6] = 'a'，即s[k] == s[j]，所以j++,k++，即k = 2,j = 7,next[j] = next[7] = k = 2,即next[7] = 2;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1; next[6] =2; nxet[7] =2;

k =2; j =7;

k = 2,j = 7,s[k] = s[2] = 'b',s[j] = s[7] = 'b'，即s[k] == s[j]，所以k++,j++，即k = 3,j = 8,nxet[j] = next[8] = k = 3,即next[8] = 3;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1; next[6] =2; nxet[7] =2; nxet[8] =3;

k =3; j =8;

 k = 3,j = 8,s[k] = s[3] = 'c',s[j] = s[8] = 'a',即s[k] != s[j]，所以k = next[k] = next[3] = 0,即k = 0，接着比较s[k]与s[j]的大小，即

k = 0,j = 8,s[k] = s[0] = 'a',s[j] = s[8] = 'a',即s[k] == s[j]，所以k++,j++，即k = 1,j = 9,next[j] = next[9] = k = 1,即next[9] = 1;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1; next[6] =2; nxet[7] =2; nxet[8] =3; next[9]= 1;

k =1; j =9;

k = 1,j = 9,s[k] = s[1] = 'a',s[j] = s[9] = 'a',即s[k] == s[j]，所以k++,j++，即k = 2,j = 10,next[j] = next[10] = k = 2,即next[10] = 2;

next[0] = -1; next[1]=0; next[2] =1; next[3] =0; next[4]=0; next[5] =1; next[6] =2; nxet[7] =2; nxet[8] =3; next[9]= 1; next[10]=2;

k =2; j =10;

最终的结果是：

序号	0	1	2	3	4	5	6	7	8	9	10
字符串	a	a	b	c	a	a	a	b	a	a	c
next值	-1	0	1	0	0	1	2	2	3	1	2

c++代码实现：

void getnext(char *s, int *next)
{
	//初始化阶段
	next[0] = -1;
	next[1] = 0;
	int k = 0, j = 1;
	while (j < strlen(s) - 1)
	{
		if (s[j] == s[k])//前串和后串相等
		{
			k++;
			j++;
			next[j] = k;
		}
		else //前串和后串不相等
		{
			if (k == 0)
			{
				j++;
				next[j] = k;
			}
			else
			{
				k = next[k];
			}
		}
	}
}

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