82 lines
3.3 KiB
C++
82 lines
3.3 KiB
C++
// Source : https://leetcode.com/problems/expression-add-operators/
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// Author : Hao Chen
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// Date : 2016-01-16
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/***************************************************************************************
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*
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* Given a string that contains only digits 0-9 and a target value, return all
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* possibilities to add binary operators (not unary) +, -, or * between the digits so
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* they evaluate to the target value.
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*
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* Examples:
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* "123", 6 -> ["1+2+3", "1*2*3"]
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* "232", 8 -> ["2*3+2", "2+3*2"]
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* "105", 5 -> ["1*0+5","10-5"]
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* "00", 0 -> ["0+0", "0-0", "0*0"]
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* "3456237490", 9191 -> []
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*
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* Credits:Special thanks to @davidtan1890 for adding this problem and creating all
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* test cases.
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***************************************************************************************/
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class Solution {
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public:
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vector<string> addOperators(string num, int target) {
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vector<string> result;
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if (num.size() == 0) return result;
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helper(num, target, result, "", 0, 0, 0, ' ');
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return result;
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}
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//DFS algorithm
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void helper(const string &num, const int target, //`num` and `target` never change
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vector<string>& result, // the array store all of the answers
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string solution, //the current potential answer.
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int idx, // the current index of `num` array
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long long val, // the current value we calculated so far
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long long prev, // the lastest value we used for calculation, which used for operation prioirty adjustment
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char preop ) // the latest "+" or "-" operation, which used for operation prioirty adjustment
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{
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if (target == val && idx == num.size()){
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result.push_back(solution);
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return;
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}
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if (idx == num.size()) return;
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string n;
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long long v=0;
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for(int i=idx; i<num.size(); i++) {
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//get rid of the number which start by "0"
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//e.g. "05" is not the case.
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if (n=="0") return;
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n = n + num[i];
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v = v*10 + num[i]-'0';
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if (solution.size()==0){
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// the first time for initialization
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helper(num, target, result, n, i+1, v, 0, ' ');
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}else{
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// '+' or '-' needn't to adjust the priority
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helper(num, target, result, solution + '+' + n, i+1, val+v, v, '+');
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helper(num, target, result, solution + '-' + n, i+1, val-v, v, '-');
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//if we meet multiply operation, we need adjust the calcualtion priority
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// e.g. if the previous value is calculated by 2+3=5,
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// then if we need to multipy 4, it is not 5*4, it is 2+3*4=2+12=24
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// we need be careful about multiply again, such as: 2+3*4*5
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if (preop=='+') {
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helper(num, target, result, solution + '*' + n, i+1, (val-prev)+prev*v, prev*v, preop);
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}else if (preop=='-'){
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helper(num, target, result, solution + '*' + n, i+1, (val+prev)-prev*v, prev*v, preop);
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}else {
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helper(num, target, result, solution + '*' + n, i+1, val*v, v, '*');
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}
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}
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}
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}
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};
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