LeetCode 282. Expression Add Operators

原题链接在这里：https://leetcode.com/problems/expression-add-operators/

题目：

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 -> []

题解：

It needs all the possible combinations. Thus it needs to use DFS.

DFS states need rest string, target, current accumlated value, the diff added last time, current combination path and res.

每次把新的num.substring按照三种方式添加到当前结果cur中，若是cur==target时, num string 也走到尾部，就把这个item结果加到res中.

从num string 中依次取出长度大于1的string来生成一个数字, cur 记录当前运算后的结果，preCurDiff用来记录最后变化. e.g. 2+3*2，即将要运算到乘以2的时候，上次循环的cur = 5, 是2 加上 3 得到的，所以preCurDiff = 3, 而要算这个乘2的时候，需要把preCurDiff先从cur中减掉，在加上新的diff. 新的 diff 是3*2=6. 

数字是可以合并出现的，比如"123", 15能返回"12+3".

若是出现 2*05 这种情况，要排除.

用Long型是防止溢出.

Time Complexity: exponential.

Space: O(n). n是num长度.

AC Java:

public class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<String>();
        dfs(num, target, 0, 0, "", res);
        return res;
    }
    
    private void dfs(String num, int target, long cur, long preCurDiff, String item, List<String> res){
        if(cur == target && num.length() == 0){ //cur加到了target 并且没有剩余的num string
            res.add(item);
            return;
        }
        
        //从头开始，每次从头取不同长度的string 作为curStr, 作为首个数字
        for(int i = 1; i<=num.length(); i++){
            String curStr = num.substring(0,i); 
            if(curStr.length() > 1 && curStr.charAt(0) == '0'){ //去掉corner case 1*05
                break;
            }
            String nextStr = num.substring(i);
            if(item.length() == 0){ //当前item为空，说明第一个数字
                dfs(nextStr, target, Long.valueOf(curStr), Long.valueOf(curStr), curStr, res);
            }else{
                dfs(nextStr, target, cur + Long.valueOf(curStr), Long.valueOf(curStr), item + "+" + curStr, res);
                dfs(nextStr, target, cur - Long.valueOf(curStr), -Long.valueOf(curStr), item + "-" + curStr, res);
                dfs(nextStr, target, cur-preCurDiff + preCurDiff*Long.valueOf(curStr),  preCurDiff*Long.valueOf(curStr), item + "*" + curStr, res);
            }
        }
    }
}