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CyC2018 2018-04-04 15:36:04 +08:00
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172
notes/Leetcode 题解.md
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@ -3223,16 +3223,16 @@ public int[] countBits(int num) {
一个栈实现：
```java
class MyQueue {
class MyQueue {
private Stack<Integer> st = new Stack();
public void push(int x) {
Stack<Integer> temp = new Stack();
while(!st.isEmpty()){
while (!st.isEmpty()) {
temp.push(st.pop());
}
st.push(x);
while(!temp.isEmpty()){
while (!temp.isEmpty()) {
st.push(temp.pop());
}
}
@ -3254,7 +3254,7 @@ class MyQueue {
两个栈实现：
```java
class MyQueue {
class MyQueue {
private Stack<Integer> in = new Stack();
private Stack<Integer> out = new Stack();
@ -3272,9 +3272,9 @@ class MyQueue {
return out.peek();
}
private void in2out(){
if(out.isEmpty()){
while(!in.isEmpty()){
private void in2out() {
if (out.isEmpty()) {
while (!in.isEmpty()) {
out.push(in.pop());
}
}
@ -3301,8 +3301,9 @@ class MyStack {
public void push(int x) {
queue.add(x);
for(int i = 1; i < queue.size(); i++){ // 翻转
queue.add(queue.remove());
int cnt = queue.size();
while (cnt-- > 1) {
queue.add(queue.poll());
}
}
@ -3341,20 +3342,14 @@ class MinStack {
public void push(int x) {
dataStack.add(x);
if(x < min) {
min = x;
}
min = Math.min(min, x);
minStack.add(min);
}
public void pop() {
dataStack.pop();
minStack.pop();
if(!minStack.isEmpty()) {
min = minStack.peek();
} else{
min = Integer.MAX_VALUE;
}
min = minStack.isEmpty() ? min = Integer.MAX_VALUE : minStack.peek();
}
public int top() {
@ -3382,17 +3377,15 @@ Output : true
```java
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if(c == '(' || c == '{' || c == '[') stack.push(c);
else{
if(stack.isEmpty()) return false;
for (char c : s.toCharArray()) {
if (c == '(' || c == '{' || c == '[') stack.push(c);
else {
if (stack.isEmpty()) return false;
char cStack = stack.pop();
if(c == ')' && cStack != '(' ||
c == ']' && cStack != '[' ||
c == '}' && cStack != '{' ) {
return false;
}
boolean b1 = c == ')' && cStack != '(';
boolean b2 = c == ']' && cStack != '[';
boolean b3 = c == '}' && cStack != '{';
if (b1 || b2 || b3) return false;
}
}
return stack.isEmpty();
@ -3729,14 +3722,9 @@ For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums sh
```java
public void moveZeroes(int[] nums) {
int n = nums.length;
int idx = 0;
for(int i = 0; i < n; i++){
if(nums[i] != 0) nums[idx++] = nums[i];
}
while(idx < n){
nums[idx++] = 0;
}
for (int num : nums) if (num != 0) nums[idx++] = num;
while (idx < nums.length) nums[idx++] = 0;
}
```
@ -3751,6 +3739,11 @@ Input: nums = [1,2,2,4]
Output: [2,3]
```
```html
Input: nums = [1,2,2,4]
Output: [2,3]
```
最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(nlogn)。本题可以以 O(n) 的时间复杂度、O(1) 空间复杂度来求解。
主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
@ -3764,32 +3757,29 @@ Output: [2,3]
```java
public int[] findErrorNums(int[] nums) {
for(int i = 0; i < nums.length; i++){
while(nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]) {
for (int i = 0; i < nums.length; i++) {
while (nums[i] != i + 1) {
if (nums[i] == nums[nums[i] - 1]) {
return new int[]{nums[nums[i] - 1], i + 1};
}
swap(nums, i, nums[i] - 1);
}
}
for(int i = 0; i < nums.length; i++){
if(i + 1 != nums[i]) {
return new int[]{nums[i], i + 1};
}
}
return null;
}
private void swap(int[] nums, int i, int j){
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
private void swap(int[] nums, int i, int j) {
int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp;
}
```
**找出数组中重复的数，数组值在 [0, n-1] 之间**
**找出数组中重复的数，数组值在 [1, n] 之间**
[Leetcode : 287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
要求不能修改数组，也不能使用额外的空间。
二分查找解法：
```java
@ -3812,18 +3802,17 @@ public int findDuplicate(int[] nums) {
```java
public int findDuplicate(int[] nums) {
int slow = nums[0], fast = nums[nums[0]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
fast = 0;
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
int slow = nums[0], fast = nums[nums[0]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
fast = 0;
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
```
@ -4014,18 +4003,20 @@ public ListNode deleteDuplicates(ListNode head) {
[Leetcode : 234. Palindrome Linked List (Easy)](https://leetcode.com/problems/palindrome-linked-list/description/)
要求以 O(1) 的空间复杂度来求解。
切成两半，把后半段反转，然后比较两半是否相等。
```java
public boolean isPalindrome(ListNode head) {
if(head == null || head.next == null) return true;
if (head == null || head.next == null) return true;
ListNode slow = head, fast = head.next;
while(fast != null && fast.next != null){
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
if(fast != null){ // 偶数节点，让 slow 指向下一个节点
if (fast != null) { // 偶数节点，让 slow 指向下一个节点
slow = slow.next;
}
@ -4035,14 +4026,14 @@ public boolean isPalindrome(ListNode head) {
return isEqual(l1, l2);
}
private void cut(ListNode head, ListNode cutNode){
while( head.next != cutNode ) head = head.next;
private void cut(ListNode head, ListNode cutNode) {
while (head.next != cutNode) head = head.next;
head.next = null;
}
private ListNode reverse(ListNode head){
private ListNode reverse(ListNode head) {
ListNode newHead = null;
while(head != null){
while (head != null) {
ListNode nextNode = head.next;
head.next = newHead;
newHead = head;
@ -4051,9 +4042,9 @@ private ListNode reverse(ListNode head){
return newHead;
}
private boolean isEqual(ListNode l1, ListNode l2){
while(l1 != null && l2 != null){
if(l1.val != l2.val) return false;
private boolean isEqual(ListNode l1, ListNode l2) {
while (l1 != null && l2 != null) {
if (l1.val != l2.val) return false;
l1 = l1.next;
l2 = l2.next;
}
@ -4307,19 +4298,13 @@ There are two left leaves in the binary tree, with values 9 and 15 respectively.
```java
public int sumOfLeftLeaves(TreeNode root) {
if(root == null) {
return 0;
}
if(isLeaf(root.left)) {
return root.left.val + sumOfLeftLeaves(root.right);
}
if(root == null) return 0;
if(isLeaf(root.left)) return root.left.val + sumOfLeftLeaves(root.right);
return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
}
private boolean isLeaf(TreeNode node){
if(node == null) {
return false;
}
if(node == null) return false;
return node.left == null && node.right == null;
}
```
@ -4376,7 +4361,7 @@ Given tree s:
/ \
1 2
Given tree t:
4
4
/ \
1 2
Return true, because t has the same structure and node values with a subtree of s.
@ -4651,11 +4636,11 @@ public int findBottomLeftValue(TreeNode root) {
### 前中后序遍历
```html
1
/ \
2 3
/ \ \
4 5 6
1
/ \
2 3
/ \ \
4 5 6
```
层次遍历顺序：[1 2 3 4 5 6]
@ -4704,14 +4689,14 @@ void dfs(TreeNode root){
```java
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) return ret;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node == null) continue;
ret.add(node.val);
if (node.right != null) stack.push(node.right);
if (node.left != null) stack.push(node.left); // 先添加右子树再添加左子树，这样是为了让左子树在栈顶
stack.push(node.right); // 先右后左，保证左子树先遍历
stack.push(node.left);
}
return ret;
}
@ -4726,14 +4711,14 @@ public List<Integer> preorderTraversal(TreeNode root) {
```java
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) return ret;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node == null) continue;
ret.add(node.val);
if (node.left != null) stack.push(node.left);
if (node.right != null) stack.push(node.right);
stack.push(node.left);
stack.push(node.right);
}
Collections.reverse(ret);
return ret;
@ -4747,11 +4732,12 @@ public List<Integer> postorderTraversal(TreeNode root) {
```java
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) return ret;
Stack<TreeNode> stack = new Stack<>();
TreeNode cur = root;
while(cur != null || !stack.isEmpty()) {
while(cur != null) { // 模拟递归栈的不断深入
stack.add(cur);
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
TreeNode node = stack.pop();

2
notes/剑指 offer 题解.md
View File

@ -90,6 +90,8 @@
在一个长度为 n 的数组里的所有数字都在 0 到 n-1 的范围内。数组中某些数字是重复的，但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中任意一个重复的数字。例如，如果输入长度为 7 的数组 {2, 3, 1, 0, 2, 5, 3}，那么对应的输出是第一个重复的数字 2。
要求复杂度为 O(N) + O(1)，时间复杂度 O(N)，空间复杂度 O(1)。因此不能使用排序的方法，也不能使用额外的标记数组。
## 解题思路
这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。