【LeetCode】282.Expression Add Operators(Hard)解题报告

本文链接：https://blog.youkuaiyun.com/haoxiaoxiaoyu/article/details/79207823

本文介绍了解决LeetCode 282题——表达式加算符的一种有效方法。该题要求给定一个仅包含0-9数字的字符串和一个目标值，返回所有可能在数字间添加二元运算符（+、- 或 *），使表达式计算结果等于目标值的所有组合。

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【LeetCode】282.Expression Add Operators(Hard)解题报告

题目地址：https://leetcode.com/problems/expression-add-operators/description/
题目描述：

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.

Examples:
”123”, 6 -> [“1+2+3”, “1*2*3”]
”232”, 8 -> [“2*3+2”, “2+3*2”]
”105”, 5 -> [“1*0+5”,”10-5”]
”00”, 0 -> [“0+0”, “0-0”, “0*0”]
”3456237490”, 9191 -> []

Solution:

//枚举，典型的backtracking
//"123" 6
//pos = 0 cur = 1 path="1"           val      pre
//      pos = 1 cur = 2 path : 1+2   3         2
//                             1-2   -1       -2
//                             1*2   1-1+1*2=2   2
//          pos = 2 cur = 3 *            2-2+2*3
//time:不知道
//space:O(n)
class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<>();
        if(num.length()==0 || num==null) return res;
        helper(res,"",num,target,0,0,0);
        return res;
    }
    public void helper(List<String> res,String path,String num,int target,int pos,long val,long pre){
        if(pos == num.length()){
            if(target == val){
                res.add(path);
                return;
            }
        }
        for(int i=pos;i<num.length();i++){
            if(i!=pos && num.charAt(pos) == '0') break;
            long cur = Long.parseLong(num.substring(pos,i+1));
            if(pos==0){
                helper(res,path+cur,num,target,i+1,cur,cur);
            }else{
                helper(res,path+"+"+cur,num,target,i+1,val+cur,cur);
                helper(res,path+"-"+cur,num,target,i+1,val-cur,-cur);
                helper(res,path+"*"+cur,num,target,i+1,val-pre+pre*cur,pre*cur);
            }
        }
    }
}