add 3-5 answers

Author: Ijiamuyu

lcof2/剑指 Offer II 003. 前 n 个数字二进制中 1 的个数/Solution.cpp

@@ -7,4 +7,17 @@ class Solution {
         }
         return f;
     }
-};
+};
+
+class Solution {
+public:
+    vector<int> countBits(int n) {
+        vector<int> ans;
+        ans.push_back(0);
+        for(int i = 1; i <= n; i++)
+        {
+            ans.push_back(ans[i & (i - 1)] +1);
+        }
+        return ans;
+    }
+};

lcof2/剑指 Offer II 004. 只出现一次的数字/Solution.cpp

@@ -12,4 +12,30 @@ class Solution {
         }
         return ans;
     }
-};
+};
+
+class Solution {
+public:
+    int singleNumber(vector<int>& nums) {
+        vector<int> array(32);
+        int ret = 0;
+        for(int i = 0; i < nums.size(); i++)
+        {
+            for(int j = 0; j < 32; j++)
+            {
+                if(nums[i] & (1 << j))
+                {
+                    array[j]++;
+                }
+            }
+        }
+        for(int i = 0; i < 32; i++)
+        {
+            if(array[i] % 3)
+            {
+                ret += ((array[i] % 3) << i);
+            }
+        }
+        return ret;
+    }
+}; 

lcof2/剑指 Offer II 005. 单词长度的最大乘积/Solution.cpp

@@ -19,4 +19,35 @@ class Solution {
         }
         return ans;
     }
+};
+
+int string_to_int(string word)
+{
+    int ret = 0;
+    for(int i = 0; i < word.size(); i++)
+    {
+        ret |= 1 << (word[i] - 'a');
+    }
+    return ret;
+}
+
+class Solution {
+public:
+    int maxProduct(vector<string>& words) {
+        int max = 0, numa = 0, numb = 0;
+        int i, j;
+        for(i = 0; i < words.size() - 1; i++)
+        {
+            numa = string_to_int(words[i]);
+            for(j = i + 1;j < words.size(); j++)
+            {
+                numb = string_to_int(words[j]);
+                if( !(numa & numb))
+                {
+                    max = max > words[i].size() * words[j].size() ? max : words[i].size() * words[j].size();
+                }
+            }
+        }
+        return max;
+    }
 };