Leetcode 282. Expression Add Operators
Leetcode 282. Expression Add Operators
triplebee
发布于 2018-03-27 16:56:33
发布于 2018-03-27 16:56:33
文章被收录于专栏：计算机视觉与深度学习基础
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 -> []给定一个字符串和数字，给字符串添加双目的“+，-，*”，求所有能让字符串表达式值为给定数字的方法。
DFS，搜的时候注意，乘法的加入会改变之前运算的优先级，所以要记录前一种运算是什么，并且将之前最近的不是乘法的部分表达式值记录下来，
这样可以结合当前的值还原出“+，-”之前的值。
最后注意可能会爆int的情形，还有前导零不算的情况，003 ！= 3
class Solution {
public:
void dfs(vector<string> &res, string &num, int target, string now, int pos, long long sum, long long sum2, char op)
{
if(num.size() == pos)
{
if(target == sum2) res.push_back(now);
return ;
}
for(int i = pos + 1; i <= num.size(); i++)
{
string tmp = num.substr(pos, i - pos);
long long tmp2 = stoll(tmp);
if(to_string(tmp2) != tmp) break;
if(op == '#')
{
dfs(res, num, target, now + tmp, i, sum, tmp2, '*');
continue;
}
if(op == '+')dfs(res, num, target, now + "*" + tmp, i, sum, (sum2 - sum) * tmp2 + sum, '*');
else if(op == '-') dfs(res, num, target, now + "*" + tmp, i, sum, sum - (sum - sum2) * tmp2, '*');
else if(op == '*') dfs(res, num, target, now + "*" + tmp, i, sum, sum + (sum2 - sum) * tmp2, '*');
dfs(res, num, target, now + "+" + tmp, i, sum2, sum2 + tmp2, '+');
dfs(res, num, target, now + "-" + tmp, i, sum2, sum2 - tmp2, '-');
}
}
vector<string> addOperators(string num, int target) {
vector<string> res;
dfs(res, num, target, "", 0, 0, 0, '#');
return res;
}
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