LeetCode496. Next Greater Element I单调栈问题

题意：

496. Next Greater Element I

Easy

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.

思路：

求某数后面第一个比它大的元素是多少。 要从后往前扫，扫描的时候维持一个单调栈。 后面的元素先入栈，进入栈底，栈底元素是最大的数。 如果当前元素比栈顶元素大，那么弹出栈顶元素。如果当前元素比栈顶元素大，那么弹出栈顶元素。如果当前元素比栈顶元素大，那么弹出栈顶元素。 栈顶元素大于当前元素的时候不再弹出，此时的栈顶元素就是当前元素后面第一个比当前元素大的数字。 如果栈为空，说明当前元素后面没有符合要求比它大的数字。

记录完当前元素的情况后，将当前元素压入栈中。 一直这样遍历nums2。 每找到一个当前元素，它的后面有比自己大的数字，就组成一个<key,value>，加入到map中。

代码：

Java：

class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { //思路见旁边的note Map<Integer,Integer> map=new HashMap<>(); Stack<Integer> stack=new Stack<>(); for(int i=nums2.length-1;i>=0;--i){//z注意是i>=,而不是简单的i>0 while(!stack.isEmpty()&&stack.peek()<=nums2[i]) stack.pop(); if(stack.isEmpty()) map.put(nums2[i],-1); else map.put(nums2[i],stack.peek()); stack.push(nums2[i]);//别忘记要入栈 } int[] res=new int[nums1.length]; for(int i=0;i<nums1.length;++i){ res[i]=map.get(nums1[i]); } return res; } }