[Algorithm]查找

一.查找的算法

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1.顺序查找

1 int Search_Seq( SeqList L, ElemType elem )
2 {
3     L.elem[0] = elem;
4     for ( int i = L.length; L.elem[0].key != L.elem[i].key; i-- );
5     return i;
6 }

2.折半查找(二分查找)

 1 int Binary_Search( SeqList L, ElemType elem )
 2 {
 3     int low = 1, high = L.length;
 4     while (low <= high)
 5     {
 6         int mid = ( low + high ) / 2;
 7         if ( elem.key == L.elem[mid].key )
 8             return mid;
 9         else if ( elem.key < L.elem[mid].key )
10             high = mid - 1;
11         else
12             low = mid + 1;
13     }
14     return 0;
15 }

3.折半查找(二分查找)递归

1 int Binary_search_RC( SeqList L, ElemType elem, int low, int high )
2 {
3     if ( low > high ) return 0;
4     int mid = ( low + high ) / 2;
5     if ( elem.key == L.elem[mid] ) return mid;
6     else if ( elem.key < L.elem[mid] )  high = mid - 1;
7     else low = mid + 1;
8     return Binary_search_RC( L, elem, low, high );
9 }

二.字符串匹配

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1.简单的模式匹配算法(朴素模式匹配算法)

 1 int Index( SString S, SString T )
 2 {
 3     int i = 1, j = 1;
 4     while ( i <= S[0] && j <= T[0] )
 5     {
 6         if ( S[i] == T[j] ) { i++; j++; }
 7         else { i = i - j + 2; j = 1; }
 8     }
 9     if ( j > T[0] ) return i - T[0];
10     return 0;
11 }

2.KMP算法

算法需要先求出模式串的next值:

 1 void get_next( SString T, int next[] )
 2 {
 3     int i = 1, j = 0;
 4     next[1] = 0;
 5     while (i<=T[0])
 6     {
 7         if (j==0||T[i]==T[j] )
 8         {
 9             i++; j++; next[i] = j;
10         }
11         else j = next[j];
12     }
13 }

也可求出改进后的nextval值:

 1 void get_nextval( SString T, int nextval[] )
 2 {
 3     int i = 1, j = 0;
 4     nextval[1] = 0;
 5     while ( i <= T[0] )
 6     {
 7         if ( j == 0 || T[i] == T[j] )
 8         {
 9             i++, j++;
10             if ( T[i] == T[j] )
11                 nextval[i] = nextval[j];
12             else
13                 nextval[i] = j;
14         }
15         else j = nextval[j];
16     }
17 }

以下是KMP算法:

 1 int KMP( SString S, SString T, int next[], int pos )
 2 {
 3     int i = pos, j = 1;
 4     while ( i<=S[0]&&j<=T[0] )
 5     {
 6         if ( j == 0 || S[i] == T[j] ) { i++; j++; }
 7         else j = next[j];
 8     }
 9     if ( j > T[0] ) return i - T[0];
10     return 0;
11 }