add 282 Expression Add Operators

Author: selfboot

DepthFirstSearch/282_ExpressionAddOperators.cpp

@@ -0,0 +1,64 @@
+/*
+ * @Author: [email protected]
+ * @Last Modified time: 2016-07-27 10:42:26
+ */
+
+class Solution {
+public:
+    /*
+    Once you can understand the solution space tree, you just get it.
+
+    Refer to:
+    https://discuss.leetcode.com/topic/24523/java-standard-backtrace-ac-solutoin-short-and-clear
+    */
+    vector<string> addOperators(string num, int target) {
+        vector<string> ans;
+        dfs_search(ans, "", num, target, 0, 0, 0);
+        return ans;
+    }
+
+    void dfs_search(vector<string> &ans, string path, const string &num,
+                    int target, int pos, long value, long pre_num){
+        /*
+        Put binary operator in pos, and then calculate the new value.
+
+        @pre_num: when process *, we need to know the previous number.
+        */
+        if(pos == num.size()){
+            if(value == target){
+                ans.push_back(path);
+            }
+            return;
+        }
+
+        for(int i=1; i+pos<=num.size(); i++){
+            string cur_str = num.substr(pos, i);
+            // Digit can not begin with 0 (01, 00, 02 are not valid), except 0 itself.
+            if(i>1 && cur_str[0] == '0')    break;
+            long cur_d = stoll(cur_str);
+            if(pos==0){
+                dfs_search(ans, cur_str, num, target, pos+i, cur_d, cur_d);
+            }
+            // All three different binary operators: +, -, *
+            else{
+                dfs_search(ans, path+"+"+cur_str, num, target, pos+i, value+cur_d, cur_d);
+                dfs_search(ans, path+"-"+cur_str, num, target, pos+i, value-cur_d, -cur_d);
+                dfs_search(ans, path+"*"+cur_str, num, target, pos+i,
+                           value-pre_num+pre_num*cur_d, cur_d*pre_num);
+            }
+        }
+    }
+};
+
+/*
+"000"
+0
+"123"
+6
+"232"
+8
+"1005"
+5
+"3456237490"
+9191
+*/

DepthFirstSearch/282_ExpressionAddOperators.py

@@ -0,0 +1,55 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+# @Author: [email protected]
+# @Last Modified time: 2016-07-27 09:43:06
+
+
+class Solution(object):
+    def addOperators(self, num, target):
+        """ Once you can understand the solution space tree, you just get it.
+
+        Refer to:
+        https://discuss.leetcode.com/topic/24523/java-standard-backtrace-ac-solutoin-short-and-clear
+        """
+        ans = []
+        self.dfs_search(ans, "", num, target, 0, 0, 0)
+        return ans
+
+    def dfs_search(self, ans, path, num, target, pos, pre_num, value):
+        """  Put binary operator in pos, and then calculate the new value.
+
+        @pre_num: when process *, we need to know the previous number.
+        """
+        if pos == len(num):
+            if value == target:
+                ans.append(path)
+            return
+
+        for i in range(pos + 1, len(num) + 1):
+            cur_str, cur_n = num[pos: i], int(num[pos: i])
+            # Digit can not begin with 0 (01, 00, 02 are not valid), except 0 itself.
+            if i > pos + 1 and num[pos] == '0':
+                break
+            if pos == 0:
+                self.dfs_search(ans, path + cur_str, num, target, i, cur_n, cur_n)
+            # All three different binary operators: +, -, *
+            else:
+                self.dfs_search(ans, path + "+" + cur_str, num,
+                                target, i, cur_n, value + cur_n)
+                self.dfs_search(ans, path + "-" + cur_str, num,
+                                target, i, -cur_n, value - cur_n)
+                self.dfs_search(ans, path + "*" + cur_str, num,
+                                target, i, pre_num * cur_n, value - pre_num + pre_num * cur_n)
+
+"""
+"000"
+0
+"123"
+6
+"232"
+8
+"1005"
+5
+"3456237490"
+9191
+"""

DepthFirstSearch/README.md

@@ -33,8 +33,11 @@ WikiPedia 里对 DFS 的说明如下：
 
 [Clone Graph](https://leetcode.com/problems/clone-graph/) 是一个图的复制问题，可以用 DFS 来遍历图中所有的顶点。
 
+# 例子，更好的理解
 
-更多阅读
+## 282. Expression Add Operators
+
+# 更多阅读
 
 [Depth-first search](https://en.wikipedia.org/wiki/Depth-first_search)