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.
"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 -> []
public class Solution {
public List<String> addOperators(String num, int target) {
List<String> res = new ArrayList<String>();
dfs(num, target, res, "", 0, 0, 0);
return res;
}
public void dfs(String num, int target, List<String> res, String path, int pos, long eval, long mult){
if(pos == num.length()){
if(eval == target)
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){
dfs(num, target, res, path + cur, i+1, cur, cur);
} else {
dfs(num, target, res, path + "+" + cur, i+1, eval + cur, cur);
dfs(num, target, res, path + "-" + cur, i+1, eval - cur, -cur);
/*
唯一需要注意的就是乘法的情况,"345" 产生出 3+4*5,
到达5的时候,eval = 7, mul 4
eval-mult 算出不包含4的值,这里是7-4=3, 4是乘号以前的算式值,4*5, 算乘法部分。
参考Basic calculator II
*/
dfs(num, target, res, path + "*" + cur, i+1, (eval-mult) + mult*cur , mult*cur);
}
}
}
}
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