Merge remote-tracking branch 'doocs/master'

# Conflicts:
#	solution/0033.Search in Rotated Sorted Array/Solution.java

Author: bluesword12350

README.md

@@ -0,0 +1,9 @@
+class Solution {
+    private final String[] M = {"", "M", "MM", "MMM"};
+    private final String[] C = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
+    private final String[] X = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
+    private final String[] I = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
+    public String intToRoman(int num) {
+        return M[num / 1000] + C[(num % 1000) / 100] + X[(num % 100) / 10] + I[num % 10];
+    }
+}

Rename folder.go

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+class Solution {
+public:
+     vector<vector<int> > threeSum(vector<int> &num) {
+        return threeSumGen(num, 0);
+    }
+    
+    vector<vector<int> > threeSumGen(vector<int> &num, int target) {
+        vector<vector<int>> allSol;
+        if(num.size()<3) return allSol;
+        sort(num.begin(),num.end());
+        
+        for(int i=0; i<num.size()-2; i++) {
+            if(i>0 && num[i]==num[i-1]) continue; // 去重
+            
+            int left=i+1, right=num.size()-1;
+            
+            while(left<right) {
+                int curSum = num[left]+num[right];
+                int curTarget = target-num[i];
+                if(curSum==curTarget) {
+                    vector<int> sol;
+                    sol.push_back(num[i]);
+                    sol.push_back(num[left]);
+                    sol.push_back(num[right]);
+                    allSol.push_back(sol);
+                    left++;
+                    right--;
+                    while(num[left]==num[left-1]) left++;
+                    while(num[right]==num[right+1]) right--;
+                }
+                else if(curSum<curTarget)
+                    left++;
+                else
+                    right--;
+            }
+        }
+        return allSol;
+    }
+    
+    
+    
+//backtrace TLE  
+    
+// 	vector<vector<int>> threeSum(vector<int>& nums) {
+// 		sort(nums.begin(), nums.end());
+// 		if (nums.size() <= 2 || nums[0] > 0) return{};
+// 		set<vector<int>> res;
+// 		for (int i = 0; i < nums.size(); i++)
+// 		{
+// 			int target = 0 - nums[i];
+// 			int m = 0, n = nums.size() - 1;
+// 			while (m < n)
+// 			{
+// 				while (true)
+// 				{
+//                     if (m >= n) break;
+// 					if (m == i)
+// 					{
+// 						m++;
+// 						continue;
+// 					}
+// 					if (n == i)
+// 					{
+// 						n--;
+// 						continue;
+// 					}
+// 					if (nums[m] + nums[n] == target) {
+// 						vector<int> s = { nums[m++], nums[i], nums[n] };
+// 						sort(s.begin(), s.end());
+//  						res.insert(s);
+// 						break;
+// 					}
+// 					if (nums[m] + nums[n] > target) n--;
+// 					if (nums[m] + nums[n] < target) m++;
+// 				}
+// 			}
+			
+// 		}
+// 		vector<vector<int>> fres(res.begin(), res.end());
+// 		return fres;
+// 	}
+};
+
+// accelerate the cin process.
+int x = []() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(NULL);
+    return 0;
+}();

solution/0008.String to Integer (atoi)/Solution.java renamed to solution/0008.String to Integer(atoi)/Solution.java

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+## 从前序与中序遍历序列构造二叉树
+
+### 问题描述
+根据一棵树的前序遍历与中序遍历构造二叉树。
+
+注意:
+你可以假设树中没有重复的元素。
+
+例如，给出
+```
+前序遍历 preorder = [3,9,20,15,7]
+中序遍历 inorder = [9,3,15,20,7]
+```
+
+返回如下的二叉树：
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+---------------
+
+### 思路
+
+利用树的前序遍历和中序遍历特性+递归实现
+
+对树进行前序遍历，每一个元素，在树的中序遍历中找到概元素；在中序遍历中，该元素的左边是它的左子树的全部元素，右边是它的右子树的全部元素，以此为递归条件，确定左右子树的范围
+
+```CPP
+/**
+ * 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) {
+        int pRight = preorder.size()-1;
+        int iRight = inorder.size()-1;
+        return build(preorder,inorder,0,pRight,0,iRight);
+    }
+    
+    
+    TreeNode* build(vector<int>& preorder,vector<int>& inorder,int pLeft,int pRight,int iLeft,int iRight){
+        if(pLeft > pRight)return NULL;
+        
+        TreeNode *node = new TreeNode(preorder[pLeft]);
+        int idx = iLeft;
+        
+        while(inorder[idx] != preorder[pLeft])idx++;
+        node->left = build(preorder,inorder,pLeft+1,pLeft+idx-iLeft,iLeft,idx-1);
+        node->right = build(preorder,inorder,pLeft+idx-iLeft+1,pRight,idx+1,iRight);  
+        return node;
+    }
+};
+
+```
+
+

solution/0008.String to Integer (atoi)/Solution.js renamed to solution/0008.String to Integer(atoi)/Solution.js

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+/**
+ * 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) {
+        int pRight = preorder.size()-1;
+        int iRight = inorder.size()-1;
+        return build(preorder,inorder,0,pRight,0,iRight);
+    }
+    
+    
+    TreeNode* build(vector<int>& preorder,vector<int>& inorder,int pLeft,int pRight,int iLeft,int iRight){
+        if(pLeft > pRight)return NULL;
+        
+        TreeNode *node = new TreeNode(preorder[pLeft]);
+        int idx = iLeft;
+        
+        while(inorder[idx] != preorder[pLeft])idx++;
+        node->left = build(preorder,inorder,pLeft+1,pLeft+idx-iLeft,iLeft,idx-1);
+        node->right = build(preorder,inorder,pLeft+idx-iLeft+1,pRight,idx+1,iRight);  
+        return node;
+    }
+};