力扣hot100做题整理（21-30）

1.搜索二维矩阵Ⅱ

当前值大于目标值，列--，小于当前值行++

class Solution {public boolean searchMatrix(int[][] matrix, int target) {int m = 0, n = matrix[0].length-1;while (m < matrix.length && n >= 0) {if (matrix[m][n] > target) {n--;} else if (matrix[m][n] < target) {m++;} else {return true;}}return false;}
}

2.相交链表

a+c+b = b+c+a。走相同的路程

public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode list1 = headA;ListNode list2 = headB;while (list1 != list2) { // 无交点null == nulllist1 = list1 != null ? list1.next : headB;list2 = list2 != null ? list2.next : headA; }return list1;}
}

3.反转链表

注意：dummy必须是null,如果new的话默认一个元素为0

class Solution {public ListNode reverseList(ListNode head) {ListNode dummy = null;while (head != null) {ListNode temp = head.next;head.next = dummy;dummy = head;head = temp;}return dummy; }
}

4.回文链表

1.快慢指针找到中间节点的前一个节点
2.翻转后面部分节点
3.同时比较值是否相等

class Solution {public boolean isPalindrome(ListNode head) {// 1.快慢指针找到中间节点前一个节点ListNode slow = head;ListNode fast = head.next;while (fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;}ListNode temp = slow.next;slow.next = null;// 2.翻转链表ListNode list2 = reverse(temp);// 3.比较判断是否回文while (head != null && list2 != null) {if (head.val == list2.val) {head = head.next;list2 = list2.next;} else {return false;}}return true;}public ListNode reverse(ListNode head) {ListNode dummy = null;while (head != null) {ListNode temp = head.next;head.next = dummy;dummy = head;head = temp;}return dummy;}
}

5.环形链表

快慢指针的应用：如果有环，其中快指针以相对速度为1接近慢指针

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

6.环形链表Ⅱ

考察快慢指针
假设相交点为y
则(x+y) * 2 = x + y + n(y+z) ==> x + y = n(y+z) ==> x = (n-1)(y+z) + z == > x = z

public class Solution {public ListNode detectCycle(ListNode head) {ListNode slow = head;ListNode fast = head;// 1.找到相遇的节点while (fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;if (slow == fast) {while (slow != head) {slow = slow.next;head = head.next;}return slow;} }return null;}
}

7.合并两个有序链表

双指针的应用
思路：构造新的链表，不断判断大小并且加入

class Solution {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode newList = new ListNode();ListNode dummy = newList;while (list1 != null && list2 != null) {if (list1.val < list2.val) {newList.next = list1;list1 = list1.next;} else {newList.next = list2;list2 = list2.next;}newList = newList.next;}if (list1 != null) newList.next = list1;if (list2 != null) newList.next = list2;return dummy.next;}
}

8.两数相加

考察携带因子，取余数等操作

class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode dummy = new ListNode();ListNode dummy_head = dummy;int carry = 0;while (l1 != null || l2 != null || carry != 0) {if(l1 != null) {carry += l1.val;l1 = l1.next;}if (l2 != null) {carry += l2.val;l2 = l2.next;}dummy.next = new ListNode(carry % 10);carry = carry / 10;dummy = dummy.next;}return dummy_head.next;}
}

9.删除链表的倒数第n个节点

虚拟头节点的使用
注意：倒数第n个节点，可能是头节点，所以需要用虚拟头节点，避免单独考虑

class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {// 1.快指针走n步ListNode dummy = new ListNode();dummy.next = head;ListNode fast = dummy;ListNode slow = dummy;while (n > 0) {fast = fast.next;n--;}while (fast.next != null) {slow = slow.next;fast = fast.next;}slow.next = slow.next.next;return dummy.next;}
}

10.两两交换链表中的节点

定义first和second指针，注意顺序first要先指向后后面

class Solution {public ListNode swapPairs(ListNode head) {ListNode dummy = new ListNode();dummy.next = head;ListNode pre = dummy;while (pre.next != null && pre.next.next != null) {ListNode first = pre.next;ListNode second = pre.next.next;pre.next = second;first.next = second.next; // 关键second.next = first;pre = first;}return dummy.next;}
}