Add comments to explain the solutions

src/grayCode/grayCode.cpp

@@ -27,12 +27,50 @@
 
 #include <stdio.h>
 #include <stdlib.h>
+#include <time.h>
 #include <iostream>
 #include <vector>
 using namespace std;
 
-
-vector<int> grayCode(int n) {
+/*
+ * I designed the following stupid algorithm base on the blow observation
+ * 
+ * I noticed I can use a `mirror-like` binary tree to figure out the gray code.
+ * 
+ * For example:
+ * 
+ *           0      
+ *        __/ \__   
+ *       0       1  
+ *      / \     / \ 
+ *     0   1   1   0
+ * So, the gray code as below: (top-down, from left to right)
+ *
+ *     0 0 0 
+ *     0 0 1
+ *     0 1 1
+ *     0 1 0
+ * 
+ *                  0
+ *            _____/ \_____
+ *           0             1
+ *        __/ \__       __/ \__
+ *       0       1     1       0
+ *      / \     / \   / \     / \
+ *     0   1   1   0 0   1   1   0
+ * 
+ * So, the gray code as below:
+ *
+ *     0 0 0 0 
+ *     0 0 0 1
+ *     0 0 1 1
+ *     0 0 1 0
+ *     0 1 1 0
+ *     0 1 1 1 
+ *     0 1 0 1
+ *     0 1 0 0
+ */
+vector<int> grayCode01(int n) {
     vector<int> v;
     //n = 1<<n;
     
@@ -56,6 +94,29 @@ vector<int> grayCode(int n) {
     return v;
 }
 
+/*
+ * Actually, there is a better way.
+ * The mathematical way is: (num >> 1) ^ num; 
+ * Please refer to http://en.wikipedia.org/wiki/Gray_code
+ */
+vector<int> grayCode02(int n) {
+    vector<int> ret;   
+    int size = 1 << n;   
+    for(int i = 0; i < size; ++i) {
+        ret.push_back((i >> 1)^i);   
+    }
+    return ret;   
+}
+
+//random invoker
+vector<int> grayCode(int n) {
+    srand(time(0));
+    if (rand()%2){
+        return grayCode01(n);
+    }
+    return grayCode02(n);
+}
+
 void printBits(int n, int len){
     for(int i=len-1; i>=0; i--) {
         if (n & (1<<i)) {
@@ -69,7 +130,7 @@ void printBits(int n, int len){
 void printVector(vector<int>& v, int bit_len)
 {
     vector<int>::iterator it;
-    
+
     for(it=v.begin(); it!=v.end(); ++it){
         //bitset<bit_len> bin(*it);
         printBits(*it, bit_len);
@@ -83,7 +144,7 @@ int main(int argc, char** argv)
 {
     int n = 2;
     if (argc>1){
-       n = atoi(argv[1]); 
+        n = atoi(argv[1]); 
     }
     vector<int> v = grayCode(n);
     printVector(v, n);

src/longestSubstringWithoutRepeatingCharacters/longestSubstringWithoutRepeatingCharacters.cpp

@@ -15,7 +15,16 @@
 #include <string>
 #include <map>
 using namespace std;
-
+/*
+ * Idea:
+ * 
+ * Using a map store each char's index.
+ * 
+ * So, we can be easy to know the when duplication and the previous duplicated char's index.
+ * 
+ * Then we can take out the previous duplicated char, and keep tracking the maxiumn length. 
+ * 
+ */
 int lengthOfLongestSubstring1(string s) {
     map<char, int> m;
     int maxLen = 0;
@@ -66,6 +75,6 @@ int main(int argc, char** argv)
         s = argv[1];
         cout << s << " : " << lengthOfLongestSubstring(s) << endl;
     }
-    
+
     return 0;
 }

src/wordSearch/wordSearch.cpp

@@ -31,13 +31,15 @@
 #include <string>
 using namespace std;
 
+//Recursive backtracking algorithm
 bool exist(vector<vector<char> > &board, string word, int idx, int row, int col, vector< vector<int> > &mask) {
     int i = row;
     int j = col;
     if (board[i][j] == word[idx] && mask[i][j]==0 ) {
-        mask[i][j]=1;
+        mask[i][j]=1; //mark the current char is matched
         if (idx+1 == word.size()) return true;
-        idx++;
+        //checking the next char in `word` through the right, left, up, down four directions in the `board`.
+        idx++; 
         if (( i+1<board.size()    && exist(board, word, idx, i+1, j, mask) ) ||
             ( i>0                 && exist(board, word, idx, i-1, j, mask) ) ||
             ( j+1<board[i].size() && exist(board, word, idx, i, j+1, mask) ) ||
@@ -45,11 +47,9 @@ bool exist(vector<vector<char> > &board, string word, int idx, int row, int col,
         {
              return true;
         }
-        mask[i][j]=0;
+        mask[i][j]=0; //cannot find any successful solution, clear the mark. (backtracking)
     }
-    //if (board[i][j] != word[idx]) {
-    //    mask[i][j]=0;
-    //}
+
     return false;
 }
 
@@ -57,11 +57,9 @@ bool exist(vector<vector<char> > &board, string word) {
     if (board.size()<=0 || word.size()<=0) return false;
     int row = board.size();
     int col = board[0].size();
-    vector< vector<int> > mask;
-    for(int i=0; i<row; i++) {
-        vector<int> v(col);
-        mask.push_back(v);
-    }
+    //using a mask to mark which char has been selected.
+    //do not use vector<bool>, it has big performance issue, could cause Time Limit Error
+    vector< vector<int> > mask(row, vector<int> col);
 
     for(int i=0; i<board.size(); i++) {
         for(int j=0; j<board[i].size(); j++){