New Problem Solution - "Remove K Digits"

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|402|[Remove K Digits](https://leetcode.com/problems/remove-k-digits/) | [C++](./algorithms/cpp/removeKDigits/RemoveKDigits.cpp)|Medium|
 |401|[Binary Watch](https://leetcode.com/problems/binary-watch/) | [C++](./algorithms/cpp/binaryWatch/BinaryWatch.cpp)|Easy|
 |400|[Nth Digit](https://leetcode.com/problems/nth-digit/) | [C++](./algorithms/cpp/nthDigit/NthDigit.cpp)|Easy|
 |399|[Evaluate Division](https://leetcode.com/problems/evaluate-division/) | [C++](./algorithms/cpp/evaluateDivision/EvaluateDivision.cpp)|Medium|

algorithms/cpp/removeKDigits/RemoveKDigits.cpp

@@ -0,0 +1,107 @@
+// Source : https://leetcode.com/problems/remove-k-digits/
+// Author : Hao Chen
+// Date   : 2016-11-11
+
+/*************************************************************************************** 
+ *
+ * Given a non-negative integer num represented as a string, remove k digits from the 
+ * number so that the new number is the smallest possible.
+ * 
+ * Note:
+ * 
+ * The length of num is less than 10002 and will be ≥ k.
+ * The given num does not contain any leading zero.
+ * 
+ * Example 1:
+ * 
+ * Input: num = "1432219", k = 3
+ * Output: "1219"
+ * Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which 
+ * is the smallest.
+ * 
+ * Example 2:
+ * 
+ * Input: num = "10200", k = 1
+ * Output: "200"
+ * Explanation: Remove the leading 1 and the number is 200. Note that the output must 
+ * not contain leading zeroes.
+ * 
+ * Example 3:
+ * 
+ * Input: num = "10", k = 2
+ * Output: "0"
+ * Explanation: Remove all the digits from the number and it is left with nothing which 
+ * is 0.
+ ***************************************************************************************/
+
+class Solution {
+public:
+    string removeKdigits_pick(string& num, int k) {
+        
+        int len = num.size();
+        string result;
+        
+        int idx = 0;
+        for (int i=0; i < len - k; i++) {
+            int min_idx = idx;
+            for (int j=min_idx; j<=i+k; j++) {
+                if (num[min_idx] > num[j])  min_idx = j;
+            }
+            
+            //don't put zero at the beginning
+            if ( !(result.empty() && num[min_idx]=='0') ){
+                result.push_back(num[min_idx]);
+            }
+            
+            //select the number started from next one, to make the order correctness.
+            idx = min_idx + 1;
+        }
+        
+        if (result.empty()) result = "0";
+        return result;
+    }
+    
+    string removeKdigits_remove(string& num, int k) {
+        if ( num.size() <= k ) return "0";
+        int left_len = num.size() - k;
+        int idx = 0;
+        for (int i=0; i<k ;i++){
+            int len = num.size();
+            for (int j=0; j<num.size()-1; j++) {
+                //if the current is bigger than next one, then revmoe the current one.
+                //In other word, we always pick the smaller one number.
+                if ( num[j] > num[j+1] ) {
+                    num.erase(j, 1);
+                    idx = j;
+                    break;
+                }
+            }
+        }
+        
+        //remove all of ZEROs at the beginning.
+        for (int i=0; i<= num.size(); i++) {
+            if (num[i] != '0' || i == num.size()) {
+                num.erase(0, i);
+                break;
+            }
+        }
+        
+        // if the digits in the array are sorted, 
+        // then, we need remove the digits at the ends.
+        if (num.size() > left_len ) {
+            num.erase(num.begin() + left_len, num.end());
+        }
+        
+        if (num.empty()) num = "0";
+        return num;
+    }
+    
+    string removeKdigits(string num, int k) {
+        srand(time(0));
+        if (rand() % 2 ) {
+            return removeKdigits_pick(num, k);
+        } else {
+            return removeKdigits_remove(num, k);
+        }
+    }
+};