Question

496. Next Greater Element I

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Description

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return _an array _ans_ of length _nums1.length_ such that _ans[i]_ is the next greater element as described above._

 

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

 

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 104
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

 

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solutions

Solution 1

class Solution:
  def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
    m = {}
    stk = []
    for v in nums2:
      while stk and stk[-1] < v:
        m[stk.pop()] = v
      stk.append(v)
    return [m.get(v, -1) for v in nums1]

class Solution {
  public int[] nextGreaterElement(int[] nums1, int[] nums2) {
    Deque<Integer> stk = new ArrayDeque<>();
    Map<Integer, Integer> m = new HashMap<>();
    for (int v : nums2) {
      while (!stk.isEmpty() && stk.peek() < v) {
        m.put(stk.pop(), v);
      }
      stk.push(v);
    }
    int n = nums1.length;
    int[] ans = new int[n];
    for (int i = 0; i < n; ++i) {
      ans[i] = m.getOrDefault(nums1[i], -1);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
    stack<int> stk;
    unordered_map<int, int> m;
    for (int& v : nums2) {
      while (!stk.empty() && stk.top() < v) {
        m[stk.top()] = v;
        stk.pop();
      }
      stk.push(v);
    }
    vector<int> ans;
    for (int& v : nums1) ans.push_back(m.count(v) ? m[v] : -1);
    return ans;
  }
};

func nextGreaterElement(nums1 []int, nums2 []int) []int {
  stk := []int{}
  m := map[int]int{}
  for _, v := range nums2 {
    for len(stk) > 0 && stk[len(stk)-1] < v {
      m[stk[len(stk)-1]] = v
      stk = stk[:len(stk)-1]
    }
    stk = append(stk, v)
  }
  var ans []int
  for _, v := range nums1 {
    val, ok := m[v]
    if !ok {
      val = -1
    }
    ans = append(ans, val)
  }
  return ans
}

function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
  const map = new Map<number, number>();
  const stack: number[] = [Infinity];
  for (const num of nums2) {
    while (num > stack[stack.length - 1]) {
      map.set(stack.pop(), num);
    }
    stack.push(num);
  }
  return nums1.map(num => map.get(num) || -1);
}

use std::collections::HashMap;

impl Solution {
  pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    let mut map = HashMap::new();
    let mut stack = Vec::new();
    for num in nums2 {
      while num > *stack.last().unwrap_or(&i32::MAX) {
        map.insert(stack.pop().unwrap(), num);
      }
      stack.push(num);
    }
    nums1
      .iter()
      .map(|n| *map.get(n).unwrap_or(&-1))
      .collect::<Vec<i32>>()
  }
}

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var nextGreaterElement = function (nums1, nums2) {
  let stk = [];
  let m = {};
  for (let v of nums2) {
    while (stk && stk[stk.length - 1] < v) {
      m[stk.pop()] = v;
    }
    stk.push(v);
  }
  return nums1.map(e => m[e] || -1);
};

Solution 2

class Solution:
  def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
    m = {}
    stk = []
    for v in nums2[::-1]:
      while stk and stk[-1] <= v:
        stk.pop()
      if stk:
        m[v] = stk[-1]
      stk.append(v)
    return [m.get(x, -1) for x in nums1]

class Solution {
  public int[] nextGreaterElement(int[] nums1, int[] nums2) {
    Deque<Integer> stk = new ArrayDeque<>();
    Map<Integer, Integer> m = new HashMap<>();
    for (int i = nums2.length - 1; i >= 0; --i) {
      while (!stk.isEmpty() && stk.peek() <= nums2[i]) {
        stk.pop();
      }
      if (!stk.isEmpty()) {
        m.put(nums2[i], stk.peek());
      }
      stk.push(nums2[i]);
    }
    int n = nums1.length;
    int[] ans = new int[n];
    for (int i = 0; i < n; ++i) {
      ans[i] = m.getOrDefault(nums1[i], -1);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
    stack<int> stk;
    unordered_map<int, int> m;
    for (int i = nums2.size() - 1; ~i; --i) {
      while (!stk.empty() && stk.top() <= nums2[i]) stk.pop();
      if (!stk.empty()) m[nums2[i]] = stk.top();
      stk.push(nums2[i]);
    }
    vector<int> ans;
    for (int& v : nums1) ans.push_back(m.count(v) ? m[v] : -1);
    return ans;
  }
};

func nextGreaterElement(nums1 []int, nums2 []int) []int {
  stk := []int{}
  m := map[int]int{}
  for i := len(nums2) - 1; i >= 0; i-- {
    for len(stk) > 0 && stk[len(stk)-1] <= nums2[i] {
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      m[nums2[i]] = stk[len(stk)-1]
    }
    stk = append(stk, nums2[i])
  }
  var ans []int
  for _, v := range nums1 {
    val, ok := m[v]
    if !ok {
      val = -1
    }
    ans = append(ans, val)
  }
  return ans
}

impl Solution {
  pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    nums1
      .iter()
      .map(|target| {
        let mut res = -1;
        for num in nums2.iter().rev() {
          if num == target {
            break;
          }
          if num > target {
            res = *num;
          }
        }
        res
      })
      .collect::<Vec<i32>>()
  }
}

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var nextGreaterElement = function (nums1, nums2) {
  let stk = [];
  let m = {};
  for (let v of nums2.reverse()) {
    while (stk && stk[stk.length - 1] <= v) {
      stk.pop();
    }
    if (stk) {
      m[v] = stk[stk.length - 1];
    }
    stk.push(v);
  }
  return nums1.map(e => m[e] || -1);
};