1、原題如下:
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 -> []
2、解題如下:
class Solution {
public:
void dfs(vector<string>& result, const string num, const int target, int pos, string os, const long calres, const char lo, const long ln)
{
if (pos == num.size() && calres == target)
result.push_back(os);
else
{
for (int i = pos + 1; i <= num.size(); i++)
{
string nos = num.substr(pos, i - pos);
long nol = stol(nos);
if (to_string(nol).size() != i - pos) continue;
dfs(result, num, target, i, os + '+' + nos, calres + nol, '+', nol);
dfs(result, num, target, i, os + '-' + nos, calres - nol, '-', nol);
dfs(result, num, target, i, os + '*' + nos, (lo == '+') ? (calres - ln + ln*nol) : (lo == '-') ? (calres + ln - ln*nol) : ln*nol, lo, ln*nol);
}
}
}
//fos:first operational string
//fol:first operational long
//nos:next operational string
//nol:next operational long
//lo:last operator
//ln:last number
vector<string> addOperators(string num, int target) {
vector<string> result;
if (!num.size()) return result;
for (int i = 1; i <= num.size(); i++)
{
string fos = num.substr(0, i);//切割出第一個操作數
long fol = stol(fos);
if (to_string(fol).size() != i) continue;
dfs(result, num, target, i, fos, fol, ' ', fol);//(point1)
}
return result;
}
};
3、總結
這道題主要用的是dfs深度檢索,遍歷所有可能出現的情況。其實我們設想,如果num.size()爲9,那就相當於有八個地方可以添加符號,我們可以選擇填或者不填,雖然每兩個數之間只能填一個操作符,但是總共填多少個並不限制,如此就會有很多種情況。point1之前首先確定第一個操作數,至於它到底有幾位,我們根據循環來逐次遍歷。然後啓動dfs遍歷功能,然後循環迭代,考慮下一個添加符號的位置在哪裏(還是逐次遍歷)操作完所有數之後就進行判斷,看是否計算結果符合,符合就插入,不符合就繼續之前的dfs~