力扣爆刷第87天之hot100五连刷21-25

力扣爆刷第87天之hot100五连刷21-25

一、240. 搜索二维矩阵 II

题目链接：https://leetcode.cn/problems/search-a-2d-matrix-ii/description/?envType=study-plan-v2&envId=top-100-liked
思路：从横向递增纵向递增的矩阵中搜索target，利用顺序的特性，从矩阵的右上角进行搜索，如果相等就返回，如果大于向下进行搜索，如果小于向左进行搜索，存在的话自然可以找到，不存在的话会一直搜索到边界才会停止，也就是当前算法还有改进的空间，可以采取早停措施。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        return dfs(matrix, 0, matrix[0].length-1, target);
    }

    boolean dfs(int[][] matrix, int i, int j, int target) {
        while(i < matrix.length && j >= 0) {
            if(matrix[i][j] == target) {
                return true;
            }
            if(matrix[i][j] > target) {
                j--;
            }else {
                i++;
            }
        }
        return false;

    }
}
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二、160. 相交链表

题目链接：https://leetcode.cn/problems/intersection-of-two-linked-lists/description/?envType=study-plan-v2&envId=top-100-liked
思路：经典题目，直接求长度，然后对齐，然后同步比较，相同即返回。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode pa = headA, pb = headB;
        int lenA = 0, lenB = 0;
        while(pa != null) {
            lenA++;
            pa = pa.next;
        }
        while(pb != null) {
            lenB++;
            pb = pb.next;
        }
        pa = headA;
        pb = headB;
        while(lenA > lenB) {
            pa = pa.next;
            lenA--;
        }
        while(lenB > lenA) {
            pb = pb.next;
            lenB--;
        }
        while(pa != null) {
            if(pa == pb) {
                return pa;
            }
            pa = pa.next;
            pb = pb.next;
        }
        return null;
    }
}
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三、206. 反转链表

题目链接：https://leetcode.cn/problems/reverse-linked-list/description/?envType=study-plan-v2&envId=top-100-liked
思路：两个指针，一个指向头结点，一个用于向后走，遍历过程中用临时变量记录临时节点，然后采用头插法即可。

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode root = new ListNode();
        ListNode p1 = root, p2 = head;
        while(p2 != null) {
            ListNode t = p2;
            p2 = p2.next;
            t.next = p1.next;
            p1.next = t;
        }
        return root.next;
    }
}
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四、234. 回文链表

题目链接：https://leetcode.cn/problems/palindrome-linked-list/description/?envType=study-plan-v2&envId=top-100-liked
思路：直接快慢指针定位中间节点，然后翻转其中一般的链表，然后遍历比较即可。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode p1 = head, p2 = head;
        while(p2 != null && p2.next != null) {
            p1 = p1.next;
            p2 = p2.next.next;
        }
        if(p2 != null) {
            p1 = p1.next;
        }
        p2 = reverse(p1);
        p1 = head;
        while(p2 != null) {
            if(p1.val != p2.val) {
                return false;
            }
            p1 = p1.next;
            p2 = p2.next;
        }
        return true;
    }

    ListNode reverse(ListNode head) {
        ListNode root = new ListNode();
        ListNode p1 = root, p2 = head;
        while(p2 != null) {
            ListNode t = p2;
            p2 = p2.next;
            t.next = p1.next;
            p1.next = t;
        }
        return root.next;
    }
}
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五、141. 环形链表

题目链接：https://leetcode.cn/problems/linked-list-cycle/description/?envType=study-plan-v2&envId=top-100-liked
思路：判断是否有环，直接快慢指针。

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