剑指 Offer 03——10
Posted 刘润森！
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@Author：Runsen
决定重新刷剑指 Offer，C++和Py版本
03. 数组中重复的数字
输入：
[2, 3, 1, 0, 2, 5, 3]
输出：2 或 3
方法： 排序 遍历、set和map
C++
class Solution {
public:
int findRepeatNumber(vector<int>& nums) {
// 排序 遍历 32 ms 22.4 MB
// sort(nums.begin(),nums.end());
// for(int i = 0; i <nums.size()-1;i++) {
// if(nums[i] == nums[i+1]){
// return nums[i];
// }
// }
// return 0;
// unordered_set 52 ms 26.7 MB
// unordered_set<int> set;
// for (int num :nums){
// if(set.count(num) == 1 ){
// return num;
// }else{
// set.insert(num);
// }
// }
// return 0;
// map 56 ms 26.8 MB
unordered_map<int,bool> map;
for(int num :nums){
if (map[num]) {
return num;
}else{
map[num] = true;
}
}
return -1;
}
};
python
class Solution:
def findRepeatNumber(self, nums: List[int]) -> int:
# nums.sort()
# for i in range(len(nums)):
# if nums[i] == nums[i+1]:
# return nums[i]
# return -1
# newset = set()
# for i in nums:
# if i in newset:
# return i
# else:
# newset.add(i)
# return -1
newmap = {}
for i in nums:
if i in newmap:
return i
else:
newmap[i] = 1
return -1
04. 二维数组中的查找
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。
方法：暴力和从右上角到左下角
C++
class Solution {
public:
bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {
// 暴力 12 ms 12.7 MB
// for(int i = 0; i < matrix.size(); i++ ){
// for (int j = 0; j <matrix[0].size();j ++) {
// if( matrix[i][j] == target) {
// return true;
// }
// }
// }
// return false;
// 从右上角到左下角 20 ms 12.8 MB
if (matrix.empty() || matrix[0].empty()) return false;
int m = matrix.size();
int row =0;
int col = matrix[0].size() -1 ;
while (row < m && col >=0){
if( matrix[row][col] == target){
return true;
}else if(matrix[row][col] > target){
col--;
}else{
row++;
}
}
return false;
}
};
python
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
# 暴力 56 ms 19 MB
# for i in range(len(matrix)):
# for j in range(len(matrix[0])):
# if matrix[i][j] == target:
# return True
# return False
# 32 ms 18.9 MB
if not matrix:
return False
rows = len(matrix)
cols = len(matrix[0])
row = 0
col = cols -1
while (row < rows) and (col >=0 ):
if matrix[row][col] == target:
return True
elif matrix[row][col] > target:
col = col - 1
else:
row = row + 1
return False
05. 替换空格
输入：s = “We are happy.”
输出：“We%20are%20happy.”
class Solution {
public:
string replaceSpace(string s) {
// new string 0 ms 6.1 MB
// string st;
// for(char c: s){
// if( c == ' '){
// st += "%20";
// }else{
// st += c;
// }
// }
// return st;
// }
// C++ 的string可变 0 ms 6 MB
int len = s.size();
int cout = 0;
for(char c: s){
if(c == ' '){
cout++;
}
}
s.resize(len + 2 * cout);
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;
}
};
Python
class Solution:
def replaceSpace(self, s: str) -> str:
# 32 ms 15.1 MB
st = ""
for i in s:
if i == ' ':
st += "%20"
else:
st += i
return st
06. 从尾到头打印链表
输入：head = [1,3,2]
输出：[2,3,1]
Ｃ＋＋
class Solution {
public:
// vector<int> reversePrint(ListNode* head) {
// // 4 ms 8.4 MB
// vector<int> vet;
// while (head){
// vet.push_back(head->val);
// head = head->next;
// }
// reverse(vet.begin(),vet.end());
// return vet;
// }
// 4 ms 10.9 MB
vector<int> reversePrint(ListNode* head) {
if (head == NULL) return {};
vector<int> pre = reversePrint(head->next);
pre.push_back(head->val);
return pre;
}
};
python
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
# 递归 120 ms 24.3 MB
# return self.reversePrint(head.next) + [head.val] if head else []
# 数组反转 48 ms 16.3 MB
if not head : return []
stack = []
while head:
stack.append(head.val)
head = head.next
return stack[::-1]
07. 重建二叉树
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
# 遍历 132 ms 87.2 MB
if len(preorder) == 0:
return None
node = TreeNode(preorder[0])
index = inorder.index(node.val)
node.left = self.buildTree(preorder[1:index + 1],inorder[:index])
node.right = self.buildTree(preorder[index + 1 :],inorder[index+1:])
return node
C++
/**
* 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 {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty() || inorder.empty()){
return NULL;
}
// 寻找根节点
TreeNode *root = new TreeNode(preorder[0]);
auto index = find(inorder.begin(),inorder.end(), preorder[0]);
vector<int> ileft(inorder.begin(),index);
vector<int> lright(index+1,inorder.end());
int len = ileft.size();
vector<int> pleft(preorder.begin()+1,preorder.begin() +1+ len);
vector<int> pright(preorder.begin() + 1+ len, preorder.end());
root->left = buildTree(pleft,ileft);
root->right = buildTree(pright,lright);
return root;
}
};
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