算法：回溯九 Plus在数字字符串中加入加号，求所有情况的和

题目

链接：https://ac.nowcoder.com/acm/contest/3286/C

牛牛给你一个只包含字符1到9的字符串S, 你现在可以在任意两个数字字符之间插入’+'号构成一个式子，现在希望你能求出所有式子的结果之和。
NiuNiu gives you a string contains only digit characters ‘1’ to ‘9’, and tell you that you can insert a ‘+’ for any number of times between any two characters to form a formula. Now he wants to know the sum of the results of all the different formulas that can be formed.

输入描述:
输入只包含一个数字字符的字符串S(1≤∣S∣≤10)
The input is a string S only contains digit characters ‘1’ to ‘9’. (1≤∣S∣≤10)

输出描述:
输出总和
Print the sum.
示例1

输入
125
输出
176
说明
可以得到插入的结果:
125(不插入)
1 + 25 =26
12 + 5= 17
1 + 2 + 5 = 8
所以答案为: 125 + 26 + 17 + 8 = 176。
All the possible formulas:
125 (insert no ‘+’)
1 + 25 = 26
12 + 5 = 17
1 + 2 + 5 = 8
So the answer is 125 + 26 + 17 + 8 = 176.

示例2

输入
9999999999
输出
12656242944

DFS深度优先回溯算法

思路：

1. 把输入的字符串input，转换为字符数组；
2. 用回溯递归的方式，拼接字符串有两种选择，要么带+加号，要么不带+加号；
3. 退出条件为，当字符数到倒数第二个数字的时候，手动拼接最后一个字符，放到结果列表中resultList。
4. 遍历结果列表中resultList中的字符串，按照+加号分割为数字字符串，把数字字符串转换为数字累加即可。

package backtracking;

import java.util.ArrayList;
import java.util.List;

public class AddPlusSymbol {

  public static void main(String[] args) {
    AddPlusSymbol obj = new AddPlusSymbol();
    String input1 = "125";
    Long result1 = obj.addPlus(input1);
    System.out.println("input > " + input1 + " ; result > " + result1);
    String input2 = "9999999999";
    Long result2 = obj.addPlus(input2);
    System.out.println("input > " + input2 + " ; result > " + result2);
  }

  public long addPlus(String input) {
    if (input == null || input.length() == 0) {
      return 0;
    }

    long result = 0;
    List<String> resultList = new ArrayList<String>();
    char[] chars = input.toCharArray();
    // dfs
    dfs(chars, resultList, "", 0);

    // loop list sum
    for (String s: resultList) {
      String[] nums = s.split("\\+");
      for (String n: nums) {
        result += Long.parseLong(n);
      }
    }

    return result;
  }

  private void dfs(char[] chars, List<String> resultList, String s, int start) {
    // exit
    if (start == chars.length - 1) {
      resultList.add(s + chars[chars.length - 1]);
      return;
    }
    s += chars[start];
    dfs(chars, resultList, s + '+', start + 1);
    dfs(chars, resultList, s, start + 1);
  }
}

代码下载

https://github.com/zgpeace/awesome-java-leetcode/blob/master/code/LeetCode/src/backtracking/AddPlusSymbol.java