面试经典150题【61-70】

61.旋转链表

本质是调换这俩

第一步：成环。第二步：找head, 第三步：断环

class Solution {public ListNode rotateRight(ListNode head, int k) {if(head == null|| k == 0)  return head;int n = 0;			   //链表的长度ListNode tail = null;  //尾节点for(ListNode p = head; p != null ; p = p.next){tail = p;n++;}k %= n;ListNode p = head;for(int i = 0; i < n - k - 1; i++)  p = p.next;   //找到链表的第n-k个节点tail.next = head;head = p.next;p.next = null;return head;  //返回新的头节点}
}

86.分隔链表

新建两个链表，一个里面的值恒小于x,一个里面的值恒大于等于x，再合并两个链表即可。

class Solution {public ListNode partition(ListNode head, int x) {// 新建两个链表ListNode smlDummy = new ListNode(0), bigDummy = new ListNode(0);// 遍历链表ListNode sml = smlDummy, big = bigDummy;while (head != null) {// 将 < x 的节点加入 sml 节点后if (head.val < x) {sml.next = head;sml = sml.next;// 将 >= x 的节点加入 big 节点后} else {big.next = head;big = big.next;}head = head.next;}// 拼接两链表sml.next = bigDummy.next;big.next = null;return smlDummy.next;}
}

104. 二叉树的最大深度

树的问题最经典的就是DFS和BFS。
DFS：

class Solution {public int maxDepth(TreeNode root) {if(root == null) return 0;return Math.max(maxDepth(root.left),maxDepth(root.right)) +1;}
}

BFS就是创建一个队列，一行一行遍历。看看能遍历几行罢了。

100.相同的树

树这里就是递归去做就行。

class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p==null &&q==null) return true;//只有一个为Null ，那肯定不一样if(p==null) return false;if(q==null) return false;//比较值和左右子树return p.val==q.val && isSameTree(p.left,q.left) && isSameTree(p.right,q.right);}
}

226.翻转二叉树

树这边还是用递归，先处理自己的逻辑，然后直接扔给左右子节点。

class Solution {public TreeNode invertTree(TreeNode root) {if(root == null) return null;TreeNode temp=root.left;root.left=root.right;root.right=temp;invertTree(root.left);invertTree(root.right);return root;}
}

101.对称二叉树

永远是最左边的和最右边的想比较，然后往里面靠近。

class Solution {public boolean isSymmetric(TreeNode root) {if(root==null) return true;return compare(root.left, root.right);}public boolean compare(TreeNode left,TreeNode right){if(left==null && right==null) return true;if(left ==null && right!=null) return false;if(left!=null && right==null) return false;if(left.val==right.val){boolean b1=compare(left.left,right.right);boolean b2=compare(right.left,left.right);return b1&&b2;}return false;}
}

105.从前序与中序遍历序列构造二叉树

根据preorder[0]去切开Inorder数组，并且得知数量后，再行切开preorder数组，最后迭代即可。

class Solution {public TreeNode buildTree(int[] preorder, int[] inorder) {if(preorder.length == 0) return null;TreeNode root=new TreeNode(preorder[0]);int i=0;for(int j=0;j<inorder.length;j++){if(inorder[j]==preorder[0]) i=j;}int[] inorder1=Arrays.copyOfRange(inorder,0,i);int[] inorder2=Arrays.copyOfRange(inorder,i+1,inorder.length);int[] preorder1=Arrays.copyOfRange(preorder,1,1+i);int[] preorder2=Arrays.copyOfRange(preorder,1+i,preorder.length);root.left=buildTree(preorder1,inorder1);root.right=buildTree(preorder2,inorder2);return root;}
}

106.从后序和中序遍历序列构造二叉树

class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {if(postorder.length == 0) return null;TreeNode root=new TreeNode(postorder[postorder.length-1]);int i=0;for(int j=0;j<inorder.length;j++){if(inorder[j]==postorder[postorder.length-1]) i=j;}int[] inorder1=Arrays.copyOfRange(inorder,0,i);int[] inorder2=Arrays.copyOfRange(inorder,i+1,inorder.length);int[] postorder1=Arrays.copyOfRange(postorder,0,i);int[] postorder2=Arrays.copyOfRange(postorder,i,postorder.length-1);root.left=buildTree(inorder1,postorder1);root.right=buildTree(inorder2,postorder2);return root;}
}

117.填充每个节点的下一个右侧节点指针II

用个BFS就行了

class Solution {public Node connect(Node root){if(root==null) return null;LinkedList<Node> queue=new LinkedList<>();queue.addLast(root);while(!queue.isEmpty()){Node temp=null;Node pre=null;int queueSize= queue.size();for(int i=0;i<queueSize;i++){temp=queue.pollFirst();if(pre !=null) pre.next=temp;pre =temp;if(temp.left!=null) queue.addLast(temp.left);if(temp.right!=null) queue.addLast(temp.right);}temp.next=null;}return root;}
}

114.二叉树展开为链表

直接先序遍历放到数组里，然后挨个取出来建立新树即可。

class Solution {public void flatten(TreeNode root) {List<TreeNode> list = new ArrayList<TreeNode>();preorderTraversal(root, list);int size = list.size();for (int i = 1; i < size; i++) {TreeNode prev = list.get(i - 1), curr = list.get(i);prev.left = null;prev.right = curr;}}public void preorderTraversal(TreeNode root, List<TreeNode> list) {if (root != null) {list.add(root);preorderTraversal(root.left, list);preorderTraversal(root.right, list);}}
}