算法刷题日志_链表

2409. 统计共同度过的日子数

class Solution {
    int days[] = new int [] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    public int countDaysTogether(String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {
        String a = arriveAlice.compareTo(arriveBob)<0?arriveBob:arriveAlice;
        String b = leaveAlice.compareTo(leaveBob)<0?leaveAlice:leaveBob;
        int x = f(a),y=f(b);
        return Math.max(y-x+1,0);
    }
    private int f(String s){
        int i = Integer.parseInt(s.substring(0,2))-1;
        int res = 0;
        for(int j = 0;j<i;j++){
            res+=days[j];
        }
        res += Integer.parseInt(s.substring(3));
        return res;
    }
}

141. 环形链表

这里我们使用快慢指针来解决这道题，当链表中有环时，那么快慢指针必定会相遇，如果没有环则不会相遇，

public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode s = head, f = head;
        while(f != null && f.next != null){ 
            s = s.next;
            f = f.next.next;
            if(s == f) return true;
        }
        return false;
    }
}

876. 链表的中间结点

这题依旧是使用快慢指针，一个走一步一个走两步，快指针停止循环的条件为当前节点和下一个节点不为空，当快指针停下时，此时slow指针的位置就是目标位置。

142. 环形链表 II

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head, slow = head;
        while (true) {
            if (fast == null || fast.next == null) return null;
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) break;
        }
        fast = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return fast;
    }
}

剑指 Offer 22. 链表中倒数第k个节点

先让快指针移动k步，然后再让快慢指针都一步一步前进，当快指针移动到空值时，此时慢指针离尾结点距离为k-1，也就是倒数第k个节点的位置了

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode former = head, latter = head;
        for(int i = 0; i < k; i++)
            former = former.next;
        while(former != null) {
            former = former.next;
            latter = latter.next;
        }
        return latter;
    }
}

链表大都是使用快慢指针来进行解题，然后根据不同题目来确定while循环结束的条件