递归实现:
class Solution {
public:
string decodeString(string s) {
int i = 0;
return decodeString(s,i);
}
private:
string decodeString(string s,int &i) {
string res;
while(i < s.size() && s[i]!=']')
{
if(!isdigit(s[i]))
res += s[i++];
else
{
int n = 0;
//获取数字
while(i < s.size() &......
这道题看了一会发现其实使用BFS会非常简单：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNo......
方法一是使用递归实现，这里借助了成员变量，完成递归函数间值传递
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int ......
BFS实现:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNo......
最开始看到这题想到用BFS实现，写完后才发现不行，如果当前行存在偶数，我下面的代码，会把下下行的结点加进去。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr)......
首先使用先序遍历的方法:
/**
* 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* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode......
使用最直接方法，先将两个单词拼接出来，再进行整体判断：
class Solution {
public:
bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
string s1,s2;
for(auto str:word1)
s1 += str;
for(auto str:word2)
s2 += str;
return s1 == s2;
}
};
运行效率:
Runtime: 4 ms, faster than 89.96% of ......
自己写的AC代码：
class Solution {
public:
int countBalls(int lowLimit, int highLimit) {
vector<int> cnt(highLimit+1,0);
int res = 0;
for(int i = lowLimit;i <= highLimit;i++)
{
cnt[calsum(i)]++;
}
for(int i = 0;i < cnt.size();i++)
{
res = max(res,cnt[i]);
}
return res;
}
private:
int calsum......
开始分析nums序列中至多只能有一个前面元素大于后面元素，否则就不是通过排序后的序列轮转得到。
class Solution {
public:
bool check(vector<int>& nums) {
int greater_cnt = 0;
for(int i = 0;i < nums.size();i++)
{
if(nums[i] > nums[(i+1)%nums.size()] )
{
if(++greater_cnt > 1)
return false;
}
}
return true;
}
};
这里解法比较简单，分析旋......
最朴素的解法：
class Solution {
public:
int sumOfUnique(vector<int>& nums) {
int cnt[101] = {0},sum = 0;
for(auto n:nums)
cnt[n]++;
for(int i=1;i < 101;i++)
{
if(cnt[i] == 1) sum += i;
}
return sum;
}
};
运行效率:
Runtime: 4 ms, faster than 66.54% of C++ online submissions for Sum of Uni......