496.Next-Greater-Element-I.md

496. Next Greater Element I

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题目地址

https://leetcode.com/problems/next-greater-element-i/

题目描述

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.

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.

代码

Approach #1 Brute Force

Complexity Analysis:

• Time complexity: O(m * n)
• Space complexity: O(m)

class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
    int[] res = new int[findNums.length];
    int j;
    for (int i = 0; i < findNums.length; i++) {
      boolean found = false;
      for (j = 0; j < nums.length; j++) {
        if (found && nums[j] > findNums[i]) {
          res[i] = nums[j];
          break;
        }
        if (nums[j] == findNums[i]) {
          found = true;
        }
      }
      if (j == nums.length) {
        res[i] = -1;
      }
    }

    return res;
  }

}

Approach #2 Better Brute Force

class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        HashMap < Integer, Integer > hash = new HashMap < > ();
        int[] res = new int[findNums.length];
        int j;
        for (int i = 0; i < nums.length; i++) {
            hash.put(nums[i], i);
        }
        for (int i = 0; i < findNums.length; i++) {
            for (j = hash.get(findNums[i]) + 1; j < nums.length; j++) {

                if (findNums[i] < nums[j]) {
                    res[i] = nums[j];
                    break;
                }
            }
            if (j == nums.length) {
                res[i] = -1;
            }
        }
        return res;
    }
}

Approach #3 Using Stack and HashMap

Complexity Analysis

• Time complexity : O(m+n). The entire nums array(of size n) is scanned only once. The stack's n elements are popped only once. The findNums array is also scanned only once.
• Space complexity : O(m+n). stack and map of size n is used. res array of size m_m_ is used, where n_n_ refers to the length of the nums array and m_m_ refers to the length of the findNums array.

class Solution {
  public int[] nextGeneraterElement(int[] findNums, int[] nums) {
    Stack<Integer> stack = new Stack<>();
    HashMap<Integer, Integer> map = new HashMap<>();
    int[] res = new int[findNums.length];
    for (int i = 0; i < nums.length; i++) {
      // 单调递减栈
      while (!stack.empty() && nums[i] > stack.peek()) {
        int small = stack.pop();
        map.put(small, nums[i]);
      }
      stack.push(nums[i]);
    }

    while (!stack.empty()) {
      map.put(stack.pop(), -1);
    }

    for (int i = 0; i < findNums.length; i++) {
      res[i] = map.get(findNums[i]);
    }

    return res;
  }
}