https://leetcode.com/problems/next-greater-element-i/#/description
You are given two arrays (without duplicates)
nums1andnums2wherenums1’s elements are subset ofnums2. Find all the next greater numbers fornums1‘s elements in the corresponding places ofnums2.The Next Greater Number of a number x in
nums1is the first greater number to its right innums2. 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.
1 public class Solution { 2 public int[] nextGreaterElement(int[] findNums, int[] nums) { 3 int len = findNums.length; 4 int[] res = new int[len]; 5 if(len > nums.length) { 6 return res; 7 } 8 for(int i=0; i<len; i++) { 9 boolean flag = false; 10 boolean k = false; 11 for(int j=0; j<nums.length; j++) { 12 if(findNums[i] == nums[j]) { 13 flag = true; 14 continue; 15 } 16 if(flag == true && nums[j] > findNums[i]) { 17 res[i] = nums[j]; 18 k = true; 19 break; 20 } 21 } 22 if(k == false) { 23 res[i] = -1; 24 } 25 } 26 return res; 27 } 28 }
1 public int[] nextGreaterElement(int[] findNums, int[] nums) { 2 Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x 3 Stack<Integer> stack = new Stack<>(); 4 for (int num : nums) { 5 while (!stack.isEmpty() && stack.peek() < num) 6 map.put(stack.pop(), num); 7 stack.push(num); 8 } 9 for (int i = 0; i < findNums.length; i++) 10 findNums[i] = map.getOrDefault(findNums[i], -1); 11 return findNums; 12 }
数组和矩阵（3）——Next Greater Element I
原文：http://www.cnblogs.com/-1307/p/6930606.html