You are given two arrays (without duplicates) nums1
and nums2
where nums1
’s elements are subset of nums2
. Find all the next greater numbers for nums1
's elements in the corresponding places of nums2
.
The Next Greater Number of a number x in nums1
is the first greater number to its right in nums2
. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
这题就想到暴力搜索,但是太慢了。于是想着要创建一个map,保存一次遍历nums2的键值对,key为元素,value为下一个较大值;这里还要用一个栈,这个我想了半天,无法直接想出来这个栈,看了题解知道这个栈,就是保存之前遍历的值中还没有找到它的下一个较大值,所以感觉自己很LOW,慢慢学习吧。
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack<int> stack_index;
unordered_map<int,int> mymap;
for(int i = 0 ; i < nums.size() ; i ++)
{
while(!stack_index.empty() && nums[i] > nums[stack_index.top()])
{
mymap[nums[stack_index.top()]] = nums[i];
stack_index.pop();
}
stack_index.push(i);
}
for(int i = 0 ; i < findNums.size() ; i++)
findNums[i] = (mymap[findNums[i]] == 0)?-1:mymap[findNums[i]];
return findNums;
}
};