唧唧复歪歪

#include #include #include using namespace std;class Solution1 {public: vector twoSum(vector &numbers, int target) { vector v; for(int i = 0; i < numbers.size(); ++i) { for(int

#include #include using namespace std;struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:

#include #include #include using namespace std;struct TreeNode { int val; TreeNode *left, *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};vector > v;class Solution {public:

感觉写的真啰嗦#include #include #include using namespace std;struct TreeLinkNode { int val; TreeLinkNode *left, *right, *next; TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}

深搜 递归#include using namespace std;struct TreeNode { int val; TreeNode *left, *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public: bool hasPathSum(Tree

仔细，仔细，仔细！#include #include #include using namespace std;struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};cla

#include #include using namespace std;class Solution {public: vector > generate(int numRows) { vector > v; for(int i = 0; i < numRows; ++i) { if(i == 0) { ve

本解法启发自http://blog.youkuaiyun.com/pickless/article/details/12034365

没有想出来，网上搜的答案。

#include #include using namespace std;class Solution {public: vector getRow(int rowIndex) { int *t = new int[rowIndex + 1]; memset(t, 0, sizeof(int) * rowIndex); t[0] = 1; for(

非递归用栈实现先序遍历

果然脑袋混沌的时候就应该去放松一下，昨天下午xiang

好题目，总结一下。1、要求时间效率O(NlogN)，就

#include #include using namespace std;struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:

快特么气成傻逼了才AC#include #include using namespace std;stack st;class Solution {public: void reverseWords(string &s) { int i = s.find_first_not_of(" "); if(i > s.size()) { s = ""; r

stack st;class Solution {public: int evalRPN(vector &tokens) { for(int i = 0; i != tokens.size(); ++i) { if(tokens[i][0] >= '0' && tokens[i][0] = '0' && tokens[i][1] <= '9') { s

链表实现的话，快慢指针