【LeetCode】241. Different Ways to Add Parentheses
2017-03-11 19:21
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题意为得出一个表达式各种组合后的结果,而且+、-、*没有优先级的差别。可以利用分治的思想来处理,遍历该表达式每次遇到操作符便将其分为左右两部分分别再次调用递归函数,再利用该处的操作符将两边得到的结果的集合联系起来,最后得到结果。
代码如下:vector<int> diffWaysToCompute(string input) {
vector<int> ans;
for(int i=0;i<input.size();i++)
{
if(input[i]=='+'||input[i]=='-'||input[i]=='*')
{
vector<int> left=diffWaysToCompute(input.substr(0,i));
vector<int> right=diffWaysToCompute(input.substr(i+1));
for(int t=0;t<left.size();t++)
{
for(int j=0;j<right.size();j++)
{
if(input[i]=='+')
{
ans.push_back(left[t]+right[j]);
}
else if(input[i]=='-')
{
ans.push_back(left[t]-right[j]);
}
else
{
ans.push_back(left[t]*right[j]);
}
}
}
}
}
if(ans.empty())
{
int sum=0;
for(int i=0;i<input.size();i++)
{
sum=sum+(input[i]-'0')*pow(10,input.size()-i);
}
ans.push_back(sum);
}
return ans;
}
代码如下:vector<int> diffWaysToCompute(string input) {
vector<int> ans;
for(int i=0;i<input.size();i++)
{
if(input[i]=='+'||input[i]=='-'||input[i]=='*')
{
vector<int> left=diffWaysToCompute(input.substr(0,i));
vector<int> right=diffWaysToCompute(input.substr(i+1));
for(int t=0;t<left.size();t++)
{
for(int j=0;j<right.size();j++)
{
if(input[i]=='+')
{
ans.push_back(left[t]+right[j]);
}
else if(input[i]=='-')
{
ans.push_back(left[t]-right[j]);
}
else
{
ans.push_back(left[t]*right[j]);
}
}
}
}
}
if(ans.empty())
{
int sum=0;
for(int i=0;i<input.size();i++)
{
sum=sum+(input[i]-'0')*pow(10,input.size()-i);
}
ans.push_back(sum);
}
return ans;
}
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