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LeetCode "Symmetric Tree"
Recursive.class Solution { public: bool isSymmetric(TreeNode *pl, TreeNode *pr) { if (!pl && !pr) return true; if ((pl && !pr)...
分类:其他好文   时间:2014-07-22 00:36:36    阅读次数:197
C# Tips: 将 VS2012 / VS2013 的.sln文件、project文件转换成 VS2010格式
原来有一些VS2013的工程文件(.sln、.csproj),使用.Net 4.0。现需要将它们转换成VS2010格式。经实验,办法如下:(1) 在Solution文件(.sln)中:把文件头部这样的行:Microsoft Visual Studio Solution File, Format Ve...
分类:其他好文   时间:2014-07-22 00:35:35    阅读次数:195
LeetCode "Plus One"
The trick is, we can work on a reversed vector - learnt from EPI.class Solution {public: vector plusOne(vector &digits) { std::reverse(digit...
分类:其他好文   时间:2014-07-22 00:33:36    阅读次数:223
LeetCode "Generate Parentheses"
DFS and using stack to validate candidates.class Solution {public: bool isValid(const string &s) { stack stk; for (int i = 0; i &...
分类:其他好文   时间:2014-07-22 00:28:34    阅读次数:217
LeetCode "Balanced Binary Tree"
Another recursion problem.class Solution {public: int getHeight(TreeNode *p) { if (!p) return 0; int hL = 1; if (p->left) h...
分类:其他好文   时间:2014-07-21 11:10:03    阅读次数:180
LeetCode "N-Queens II"
Classic recursion\pruning problem. We can use O(n) space: A[i] = j means [i,j] is occupied.class Solution {public: int ret; bool isValid(int *A,...
分类:其他好文   时间:2014-07-21 11:08:20    阅读次数:240
LeetCode "Merge Two Sorted Lists"
Simply care about the boundary cases:class Solution {public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if (l1 && !l2) return l1;...
分类:其他好文   时间:2014-07-21 11:07:04    阅读次数:206
LeetCode "Maximum Subarray"
Very classic problem. You can brush up your DP and Searching skills.DP:class Solution {public: int maxSubArray(int A[], int n) { // dp[i + 1...
分类:其他好文   时间:2014-07-21 11:06:21    阅读次数:216
逆转字符串leetcode
public class Solution { public String reverseWords(String s) { String ans=reverse(s); String s2[]=ans.split("\\s+"); ...
分类:其他好文   时间:2014-07-21 08:20:33    阅读次数:203
【leetcode】Sum Root to leaf Numbers
简单的二叉树的先根遍历模板的应用 class Solution: # @param root, a tree node # @return an integer def hehe(self, num, root): #再原来的基础上*10,再加上当前的root.val num = num * 10 + root.val ...
分类:其他好文   时间:2014-07-20 22:45:33    阅读次数:299
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