LeetCode282. Expression Add Operators

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or *between the digits so they evaluate to the target value.

Example 1:

Input: "123", target = 6
Output: ["1+2+3", "1*2*3"] 
num =

Example 2:

Input: "232", target = 8
Output: ["2*3+2", "2+3*2"]num =

Example 3:

Input: "105", target = 5
Output: ["1*0+5","10-5"]num =

Example 4:

Input: "00", target = 0
Output: ["0+0", "0-0", "0*0"]
num =

Example 5:

Input: "3456237490", target = 9191
Output: []num =

分析

解法是利用递归回溯来遍历全部的可能，可是要注意一些边界情形。

public class Solution { public List<String> addOperators(String num, int target) { List<String> rst = new ArrayList<String>(); if(num == null || num.length() == 0) return rst; helper(rst, "", num, target, 0, 0, 0); return rst; }
　　// eval记录当前计算结果，multed计算上次计算变化的部分，在选择乘法时会用到这个 public void helper(List<String> rst, String path, String num, int target, int pos, long eval, long multed){ if(pos == num.length()){ if(target == eval) rst.add(path); return; } for(int i = pos; i < num.length(); i++){ if(i != pos && num.charAt(pos) == '0') break;　　// 抛弃以0开始的数字 long cur = Long.parseLong(num.substring(pos, i + 1)); if(pos == 0){ helper(rst, path + cur, num, target, i + 1, cur, cur); // 起始数字特殊处理 } else{ helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur);　　// 对当前数字cur选择加上以前的部分 helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur);　　// 选择减
　　　　　　　　// 选择乘法要特殊处理，减去上次变化的部分，将这个变化的部分乘以当前的数字再加上去 helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur ); } } } }