链表 2.2

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)。

#include <iostream>
#include <fstream>
#include <assert.h>

using namespace std;

struct Node {
    Node(): val(0), next(0) {}
    Node( int aval ): val(aval), next(0) {}
    int val;
    Node *next;
};

Node* findLastKth( Node *node, int k );

int main( int argc, char *argv[] ) {
    string data_file = "./2.2.txt";
    ifstream ifile( data_file.c_str(), ios::in );
    if( !ifile.is_open() ) {
        fprintf( stderr, "cannot open file: %s
", data_file.c_str() );
        return -1;
    }
    int n = 0, k = 0;
    while( ifile >>n >>k ) {
        assert( n >= 0 );
        Node guard, *node = &guard;
        for( int i = 0; i < n; ++i ) {
            node->next = new Node();
            node = node->next;
            ifile >>node->val;
            cout <<node->val <<" ";
        }
        cout <<endl;
        node = findLastKth( guard.next, k );
        if( node ) {
            printf( "last %dth node is: %d
", k, node->val );
        } else {
            printf( "last %dth node is: null", k );
        }
        node = guard.next;
        while( node ) {
            Node *next = node->next;
            delete node;
            node = next;
        }
        cout <<endl <<endl;
    }
    ifile.close();
    return 0;
}

Node* findLastKth( Node *node, int k ) {
    Node *slow = node, *fast = node;
    for( int i = 0; i < k; ++i ) {
        if( !fast ) { return 0; }
        fast = fast->next;
    }
    while( fast ) {
        slow = slow->next;
        fast = fast->next;
    }
    return slow;
}

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