282. Expression Add Operators 加减乘的全部路径
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: num =
"123", target = 6
Output: ["1+2+3", "1*2*3"]
Example 2:
Input: num =
"232", target = 8
Output: ["2*3+2", "2+3*2"]
Example 3:
Input: num =
"105", target = 5
Output: ["1*0+5","10-5"]
Example 4:
Input: num =
"00", target = 0
Output: ["0+0", "0-0", "0*0"]
Example 5:
Input: num =
"3456237490", target = 9191
Output: []
没想到的思路:
加个String path,当做参数之后 直接往结果里面加
long sum = sign? pre+current:pre-current; boolean类型的变量直接这么用
加个long pre,long current,boolean sign就行了。因为要处理不同符号的运算,这样更方便
这里先保存着pre,current*number,之后加减的时候再调用
一位接一位的时候需要。没有pos了这次,因为s = s.substring(i)不断缩短了
运算结果都是long类型
num.substring(0,i)和num.substring(i)是不一样的
退出的首要条件是s.length()==0,然后才是(int)sum == target,sum相等了就退出,不是cur相等了就退出
number * cur接着用之前的参数,cur,因为cur就是之前的number
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { List<String> results = new ArrayList<String>(); public List<String> addOperators(String s, int target) { //cc if (s == null || s == "") return results; //对于s的每一位进行for循环 for (int i = 1; i <= s.length(); i++) { if (i >= 2 && s.charAt(0) == '0') continue; dfs(s.substring(i), s.substring(0, i), 0, Long.parseLong(s.substring(0, i)), target, true); } //返回结果 return results; } public void dfs(String s, String path, long prev, long cur, int target, boolean operator) { //先定义好operator long sum = operator ? prev + cur : prev - cur; //exit if (s.length() == 0) { if (sum == (long) target) { results.add(path); return ; } } //对于s的每一位进行for循环 for (int i = 1; i <= s.length(); i++) { if (i >= 2 && s.charAt(0) == '0') continue; long number = Long.parseLong(s.substring(0, i)); dfs(s.substring(i), path + '+' + s.substring(0, i), sum, number, target, true); dfs(s.substring(i), path + '-' + s.substring(0, i), sum, number, target, false); dfs(s.substring(i), path + '*' + s.substring(0, i), prev, number * cur, target, operator); } } }