【剑指Offer】注意程序的健壮性

面试题22. 链表中倒数第k个节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        if(head == null || k == 0){
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && k-- > 0){
            fast = fast.next;
        }
        if(k > 0){
            return null;
        }
        while(fast != null && slow != null){
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}

 顺时针转圈打印矩阵！(tr,tc) 在左上角 (dr,dc)在右下角 表示这一圈打印的范围 下一圈 即 tr++tc++dr--dc--

打印一圈有三种情况：一种是完整的圈【依次打印顶边、右边、底边、左边】，一种是只有一行（tr==dr，需要从tc打印到dc），一种是只有一列(tc==dc，需要从tr打印到dr)

class Solution {
    private int index;
    private int[] ans;
    public int[] spiralOrder(int[][] matrix) {
        if(matrix == null || matrix.length == 0){
            return new int[]{};
        }
        int tr = 0, tc= 0, dr = matrix.length-1, dc = matrix[0].length-1;
        ans =new int[(dr+1)*(dc+1)];
        index = 0;
        while(tr <= dr && tc <= dc){
            process(matrix, tr++, tc++, dr--, dc--);
        }
        return ans;
    }
    public void process(int[][] matrix, int tr, int tc, int dr, int dc){
        if(tr == dr){//只有一行
            for(int i = tc; i <= dc; i++){
                ans[index++] = matrix[tr][i];
            }
        }
        else if(tc == dc){//只有一列
            for(int i = tr; i <= dr; i++){
                ans[index++] = matrix[i][tc];
            }
        }
        else{//一般情况
            int curC = tc;
            int curR = tr;
            while(curC != dc){//打印顶边
                ans[index++] = matrix[tr][curC++];
            }
            while(curR != dr){//打印右边
                ans[index++] = matrix[curR++][dc];
            }
            while(curC != tc){//打印底边
                ans[index++] = matrix[dr][curC--];
            }
            while(curR != tr){//打印左边
                ans[index++] = matrix[curR--][tc];
            }
        }
    }
}