实现一个算法,找出单向链表中倒数第k个结点。
分析:使用相差k个位置的两个指针,以相同的速度遍历链表,当快指针为空时,慢指针刚好指向链表的倒数第k个结点。时间复杂度O(n),空间复杂度O(1)。
1 #include <iostream> 2 #include <fstream> 3 #include <assert.h> 4 5 using namespace std; 6 7 struct Node { 8 Node(): val(0), next(0) {} 9 Node( int aval ): val(aval), next(0) {} 10 int val; 11 Node *next; 12 }; 13 14 Node* findLastKth( Node *node, int k ); 15 16 int main( int argc, char *argv[] ) { 17 string data_file = "./2.2.txt"; 18 ifstream ifile( data_file.c_str(), ios::in ); 19 if( !ifile.is_open() ) { 20 fprintf( stderr, "cannot open file: %s\n", data_file.c_str() ); 21 return -1; 22 } 23 int n = 0, k = 0; 24 while( ifile >>n >>k ) { 25 assert( n >= 0 ); 26 Node guard, *node = &guard; 27 for( int i = 0; i < n; ++i ) { 28 node->next = new Node(); 29 node = node->next; 30 ifile >>node->val; 31 cout <<node->val <<" "; 32 } 33 cout <<endl; 34 node = findLastKth( guard.next, k ); 35 if( node ) { 36 printf( "last %dth node is: %d\n", k, node->val ); 37 } else { 38 printf( "last %dth node is: null", k ); 39 } 40 node = guard.next; 41 while( node ) { 42 Node *next = node->next; 43 delete node; 44 node = next; 45 } 46 cout <<endl <<endl; 47 } 48 ifile.close(); 49 return 0; 50 } 51 52 Node* findLastKth( Node *node, int k ) { 53 Node *slow = node, *fast = node; 54 for( int i = 0; i < k; ++i ) { 55 if( !fast ) { return 0; } 56 fast = fast->next; 57 } 58 while( fast ) { 59 slow = slow->next; 60 fast = fast->next; 61 } 62 return slow; 63 }
测试文件
0 0
0 1
1 0
1
1 1
1
1 2
1
2 0
1 3
2 1
1 3
2 2
1 3
2 3
1 3
7 1
14 54 96 23 45 12 45
7 2
14 54 96 23 45 12 45
7 7
14 54 96 23 45 12 45
原文:http://www.cnblogs.com/moderate-fish/p/3980126.html