[LeetCode]Expression Add Operators

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.

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 -> []

分析

这道题首先想到的DFS, 递归结束时检查总值是否等于target，是的话把当前结果加入list里。

关键点是对于乘法，怎么处理当前总和的问题，这里用的方法是存一个prevVal，用来表示上一步的数，遇到乘号，就先在当前总和基础上减去上一步的值，然后再加上上一步的值与当前数的乘积。

另外有些小地方要注意，比如遇到‘0’这样的我们只考虑单独的数字0，不考虑其他以‘0’开头的字符串。

复杂度

time: O(4^n), space: O(n)

代码

public class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<String>();
        dfs(res, num, target, 0, 0, 0, "");
        return res;
    }
    
    public void dfs(List<String> res, String num, int target, int start, long currVal, long prevVal, String currExpr) {
        if (start == num.length()) {
            if (currVal == target) {
                res.add(currExpr);
            }
            return;
        }
        
        for (int i = start + 1; i <= num.length(); i++) {
            String expr = num.substring(start, i);
            long val = Long.parseLong(expr);
            
            // 对于'0'开头的非单独'0'字符串，直接跳出
            if (num.charAt(start) == '0' && i != start + 1) {
                break;
            }
            
            // 最开始的当前表达式只是加个数字
            if (start == 0) {
                dfs(res, num, target, i, val, val, expr);
            } else {
                dfs(res, num, target, i, currVal + val, val, currExpr + "+" + expr);
                dfs(res, num, target, i, currVal - val, -val, currExpr + "-" + expr);
                dfs(res, num, target, i, currVal - prevVal + prevVal * val, prevVal * val, currExpr + "*" + expr);
            }
        }
    }
}