用遍歷的辦法,挨個比,效率比較低:
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
int n=findNums.size();
int m=nums.size();
vector<int> flag;
vector<int> re;
vector<int> s;
for(int i=0;i<n;i++)
{
flag.push_back(0);
re.push_back(-2);
s.push_back(findNums[i]);
}
for(int j=m-1;j>=0;j--)
{
for(int i=0;i<n;i++)
{
if(flag[i]==1)
continue; //continue和break啊......break是for就不循環了,continue是隻跳過當前循環......
if(nums[j]==findNums[i])
{
flag[i]=1;
if(s[i]==findNums[i])
{
re[i]=-1;
}
else
{
re[i]=s[i];
}
}
else
{
if(nums[j]>findNums[i])
{
s[i]=nums[j];
}
}
}
}
return re;
}
};
leetcode上答案的一種,先用棧做一個哈希表,做出來所有元素的next_greater的表,之後查表就好:
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack<int> s;
unordered_map<int, int> m;
for (int n : nums) {
while (s.size() && s.top() < n) {
m[s.top()] = n;
s.pop();
}
s.push(n);
}
vector<int> ans;
for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1); //由於是hash表,不會有重複,所以m.count(n)(數unordered_map<int,int>前面的int中有幾個n)只會有0,1兩種結果
return ans;
}
};