LeetCode496 Next Greater Element I

文章目录
  1. 1. 描述
  2. 2. 样例
  3. 3. 思路
  4. 4. 代码

描述

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.

Note:

  1. All elements in nums1 and nums2 are unique.
  2. The length of both nums1 and nums2 would not exceed 1000.

样例

Example 1:

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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:

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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.

思路

找出nums1中的元素在nums2中对应位置的右边且比它大的第一个数,不存在则输出-1。

单调栈的应用。由于nums1是nums2的子集,所以先预处理出nums2中的所有答案再查询。

栈中的元素保持单调递减,若当前访问的元素x大于栈顶的元素sta.top(),则栈顶元素出栈并记录答案(即sta.top()右边第一个比它大的数为x)

代码

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class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
unordered_map<int, int> hs;
stack<int> sta;
for (int& x: nums) {
while (!sta.empty() && sta.top() < x) {
hs[sta.top()] = x;
sta.pop();
}
sta.push(x);
}
for (int& x: findNums) {
if (!hs.count(x)) x = -1;
else x = hs[x];
}
return findNums;
}
};
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