(简单)剑指 Offer 03. 数组中重复的数字
找出数组中重复的数字。
在一个长度为 n 的数组 nums 里的所有数字都在 0~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重复了,也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。
示例 1:
输入:
[2, 3, 1, 0, 2, 5, 3]
输出:2 或 3
限制:
2 <= n <= 100000
- 题解:
class Solution {
public:int findRepeatNumber(vector<int>& nums) {int i = 0;while(i < nums.size()){if (nums[i] == i){i++;continue;}if (nums[i] == nums[nums[i]])return nums[i];swap(nums[i], nums[nums[i]]);}return -1;}
};
(中等)剑指 Offer 04. 二维数组中的查找
减 的顺序排序。请完成一个高效的函数,输入这样的一个二维数组和一个整数,判断数组中是否含有该整数。
示例:
现有矩阵 matrix 如下:
[[1, 4, 7, 11, 15],[2, 5, 8, 12, 19],[3, 6, 9, 16, 22],[10, 13, 14, 17, 24],[18, 21, 23, 26, 30]
]
给定 target = 5
,返回 true
。
给定 target = 20
,返回 false
。
限制:
0 <= n <= 1000
0 <= m <= 1000
- 题解:
c++
class Solution {
public:bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {if (matrix.size() <= 0)return false;int i = matrix.size() - 1;int j = 0;while (i >= 0 && j < matrix[0].size()){if (target < matrix[i][j])i--;else if (target > matrix[i][j])j++;else return true;};return false;}
};
(简单)剑指 Offer 05. 替换空格
请实现一个函数,把字符串 s 中的每个空格替换成"%20"。
示例 1:
输入:s = "We are happy."
输出:"We%20are%20happy."
限制:
0 <= s 的长度 <= 10000
- 题解:
C++
class Solution {
public:string replaceSpace(string s) {int count = 0;for(auto ch: s)if (ch == ' ')count++;int len = s.size();s.resize(len + 2 * count);for (int i = len - 1, j = s.size() - 1; i < j; i--, j--)if (s[i] != ' ')s[j] = s[i];else{s[j] = '0';s[j - 1] = '2';s[j - 2] = '%';j -= 2;}return s;}
};
Golang
func replaceSpace(s string) string {ss := []byte(s)length := len(ss)count := 0for _, ch := range ss {if ch == ' ' {count++}}cmp := make([]byte, 2 * count)ss = append(ss, cmp...)for i, j := length - 1, len(ss) - 1; i < j; i, j = i - 1, j - 1 {if ss[i] != ' ' {ss[j] = ss[i]} else {ss[j] = '0'ss[j - 1] = '2'ss[j - 2] = '%'j -= 2 }}return string(ss)
}
(简单)剑指 Offer 06. 从尾到头打印链表
输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。
示例 1:
输入:head = [1,3,2]
输出:[2,3,1]
限制:
0 <= 链表长度 <= 10000
- 题解:
C++
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:vector<int> reversePrint(ListNode* head) {std::vector<int> res;while(head){res.push_back(head->val);head = head->next;}int n = res.size();for (int i = 0; i < (n >> 1); i++)swap(res[i], res[n - 1 -i]);return res;}
};
Golang
/*** Definition for singly-linked list.* type ListNode struct {* Val int* Next *ListNode* }*/
func reversePrint(head *ListNode) []int {if head == nil {return make([]int, 0)}var arrys = make([]int, 0)for head != nil {arrys = append(arrys, head.Val)head = head.Next}length := len(arrys)for i := 0; i <= (length - 1) / 2; i++ {arrys[i], arrys[length - 1 - i] = arrys[length - 1 - i], arrys[i]}return arrys
}
(中等)剑指 Offer 07. 重建二叉树
输入某二叉树的前序遍历和中序遍历的结果,请构建该二叉树并返回其根节点。
假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
示例 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
限制:
0 <= 节点个数 <= 5000
- 题解:
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
private:std::unordered_map<int, int> m_index;TreeNode* _buildTree(std::vector<int>& preorder, int preLeft, int preRight,std::vector<int>& inorder , int inLeft , int inRight){if (preLeft > preRight)return nullptr;int rootValue = preorder[preLeft];TreeNode* root = new TreeNode(rootValue);// root node value in inorder.int index = m_index[rootValue];// length of left subtree.int length = index - inLeft; // (index -1) - inLeft + 1 root->left = _buildTree(preorder, preLeft + 1, (preLeft + 1) + (length - 1),inorder, inLeft, index - 1);root->right = _buildTree(preorder, (preLeft + 1) + length, preRight,inorder, index + 1, inRight);return root;}
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int n = preorder.size();for (int i = 0; i < n; i++)m_index[inorder[i]] = i;return _buildTree(preorder, 0, n -1, inorder , 0, n -1);}
};
(简单)剑指 Offer 09. 用两个栈实现队列
用两个栈实现一个队列。队列的声明如下,请实现它的两个函数 appendTail
和 deleteHead
,分别完成在队列尾部插入整数和在队列头部删除整数的功能。(若队列中没有元素,deleteHead
操作返回 -1
)
示例 1:
输入:
["CQueue","appendTail","deleteHead","deleteHead","deleteHead"]
[[],[3],[],[],[]]
输出:[null,null,3,-1,-1]
示例 2:
输入:
["CQueue","deleteHead","appendTail","appendTail","deleteHead","deleteHead"]
[[],[],[5],[2],[],[]]
输出:[null,-1,null,null,5,2]
提示:
-
1 <= values <= 10000
-
最多会对
appendTail
、deleteHead
进行10000
次调用 -
题解:
class CQueue {
private:std::stack<int> m_stackA, m_stackB;void moveA2B() {while(!m_stackA.empty()){m_stackB.push(m_stackA.top());m_stackA.pop();}}
public:CQueue() {}void appendTail(int value) {m_stackA.push(value);}int deleteHead() {if (m_stackB.empty()) {if (m_stackA.empty())return -1;moveA2B();}int result = m_stackB.top();m_stackB.pop();return result;}
};/*** Your CQueue object will be instantiated and called as such:* CQueue* obj = new CQueue();* obj->appendTail(value);* int param_2 = obj->deleteHead();*/
(简单)剑指 Offer 10- I. 斐波那契数列
写一个函数,输入 n
,求斐波那契(Fibonacci)数列的第 n
项(即 F(N)
)。斐波那契数列的定义如下:
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
斐波那契数列由 0 和 1 开始,之后的斐波那契数就是由之前的两数相加而得出。
答案需要取模 1e9+7(1000000007),如计算初始结果为:1000000008,请返回 1。
示例 1:
输入:n = 2
输出:1
示例 2:
输入:n = 5
输出:5
提示:
0 <= n <= 100
- 题解:
class Solution {
public:int fib(int n) {if (n < 2)return n;int a = 0, b = 1, sum = 0;for (int i = 2; i <= n ; i++){sum = (a + b) % 1000000007;a = b;b = sum;}return sum ;}
};
(简单)剑指 Offer 10- II. 青蛙跳台阶问题
一只青蛙一次可以跳上1级台阶,也可以跳上2级台阶。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
答案需要取模 1e9+7(1000000007),如计算初始结果为:1000000008,请返回 1。
示例 1:
输入:n = 2
输出:2
示例 2:
输入:n = 7
输出:21
示例 3:
输入:n = 0
输出:1
提示:
0 <= n <= 100
- 题解:
class Solution {
public:int numWays(int n) {if (n == 0) return 1;if (n == 1)return 1;int a = 1, b = 1, r = 0;for (int i = 2; i <= n; i++){r = (a + b)%1000000007;a = b;b = r;}return r;}
};
(简单)剑指 Offer 11. 旋转数组的最小数字
把一个数组最开始的若干个元素搬到数组的末尾,我们称之为数组的旋转。
给你一个可能存在 重复 元素值的数组 numbers
,它原来是一个升序排列的数组,并按上述情形进行了一次旋转。请返回旋转数组的最小元素。例如,数组 [3,4,5,1,2]
为 [1,2,3,4,5]
的一次旋转,该数组的最小值为 1。
注意,数组 [a[0], a[1], a[2], ..., a[n-1]]
旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], ..., a[n-2]]
。
示例 1:
输入:numbers = [3,4,5,1,2]
输出:1
示例 2:
输入:numbers = [2,2,2,0,1]
输出:0
提示:
-
n == numbers.length
-
1 <= n <= 5000
-
-5000 <= numbers[i] <= 5000
-
numbers
原来是一个升序排序的数组,并进行了1
至n
次旋转 -
题解:
class Solution {
public:int minArray(vector<int>& numbers) {int low = 0;int high = numbers.size() - 1;while (low < high){int pivot = low + ((high - low) >> 1);if (numbers[pivot] < numbers[high])high = pivot;else if (numbers[pivot] > numbers[high])low = pivot + 1;else high--;}return numbers[low];}
};
(中等)剑指 Offer 12. 矩阵中的路径
给定一个 m x n
二维字符网格 board
和一个字符串单词 word
。如果 word
存在于网格中,返回 true
;否则,返回 false
。
单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
例如,在下面的 3×4 的矩阵中包含单词 “ABCCED”(单词中的字母已标出)。
示例 1:
输入:board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出:true
示例 2:
输入:board = [["a","b"],["c","d"]], word = "abcd"
输出:false
提示:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board
和word
仅由大小写英文字母组成
题解:
class Solution {
private:int m_columns, m_rows;bool dfs(vector<vector<char>>& board, string word, int i, int j, int index){if (i >= m_rows ||j >= m_columns ||i < 0 ||j < 0 ||board[i][j] != word[index])return false;if (index == word.size() - 1)return true;board[i][j] = '\0';bool result = dfs(board, word, i + 1, j, index + 1) || dfs(board, word, i - 1, j, index + 1) ||dfs(board, word, i, j + 1, index + 1) || dfs(board, word, i, j - 1, index + 1);board[i][j] = word[index];return result;}public:bool exist(vector<vector<char>>& board, string word) {m_rows = board.size();m_columns = board[0].size();for (int i = 0; i < m_rows; i++)for (int j = 0; j < m_columns; j++)if (dfs(board, word, i, j, 0))return true;return false;}
};