您的位置:首页 > 其它

LeetCode 2016.3.7 282,241

2016-03-07 22:36 417 查看
282 Expression Add Operators

class Solution {
public:
vector<string> ans;
long long int strtoNum(string str)
{
long long int p=1;
if (str[0]=='-')
{
p=-1;
str=str.substr(1);
}
long long int ans=0;
for(long long int i=0;i<str.size();i++)
{
ans= ans*10+str[i]-'0';
}
return ans*p;
}
vector<string> addOperators(string num, long long int target) {
dfs(num,target,0,0,"");
return ans;
}
void dfs(string num,long long int target,long long int currentSum, long long int preVal,string currentExp)
{
if (num.size()==0)
{
if (currentSum==target)
{
ans.push_back(currentExp);
}
return ;
}
for(long long int i=1;i<=num.size();i++)
{
string left=num.substr(0,i);
string right=num.substr(i);
long long int currentVal=strtoNum(left);
if (num[0]=='0' && left.size()>1) break;
if (currentExp.size()==0)
{
dfs(right,target,currentVal,currentVal,left);
}
else
{
dfs(right,target,currentSum+currentVal,currentVal,currentExp+"+"+left);
dfs(right,target,currentSum-currentVal,-currentVal,currentExp+"-"+left);
dfs(right,target,currentSum-preVal+preVal*currentVal,currentVal*preVal,currentExp+"*"+left);
}
}
return ;
}
};


241 Different Ways to Add Parentheses

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> ans;
int len=input.size();
for(int i=0;i<len;i++)
{
if (input[i]=='+' || input[i]=='-' || input[i]=='*')
{
vector<int> left = diffWaysToCompute(input.substr(0,i));
vector<int> right = diffWaysToCompute(input.substr(i+1));
int lenleft=left.size();
int lenright=right.size();
for(int j=0;j<lenleft;j++)
{
for(int k=0;k<lenright;k++)
{
if (input[i]=='+')
ans.push_back(left[j]+right[k]);
else
if (input[i]=='-')
ans.push_back(left[j]-right[k]);
else
if (input[i]=='*')
ans.push_back(left[j]*right[k]);
}
}
}
}
if(ans.empty())
ans.push_back(atoi(input.c_str()));
return ans;
}
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: