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摘要：
use dynamic programming to exchange time with space. 1 public class Solution { 2 public int uniquePaths(int m, int n) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 int[][] mytable = new... 阅读全文
摘要：
very easy recursion1 public class Solution {2 public int maxDepth(TreeNode root) {3 // IMPORTANT: Please reset any member data you declared, as4 // the same Solution instance will be reused for each test case.5 if(root == null)6 return 0;7 return Math.... 阅读全文
摘要：
特殊值可先判断并返回 很重要本题是通过左移代替乘法产生不同的除数， 然后做减法得到解。 1 public class Solution { 2 public int divide(int dividend, int divisor) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(dividend == 0) 6 ... 阅读全文
摘要：
当数组为 1，2，3，4，.. i，.. n时，基于以下原则的BST建树具有唯一性：以i为根节点的树，其左子树由[0, i-1]构成， 其右子树由[i+1, n]构成。再由递归便可得:public static int numTrees(int n) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. int[] myarray = new int[n+... 阅读全文
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