[Leetcode] 496. Next Greater Element I 解题报告

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本文介绍了一种高效算法，用于找出数组中每个元素右侧的第一个更大的数，并应用于两个给定数组的问题解决。通过使用栈和哈希表，实现了O(n)的时间复杂度。

题目：

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.

Note:

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

思路：

我们定义一个哈希表，用来记录num右边的第一个greater number。怎么建立哈希表呢？答案是对nums进行扫描，如果栈为空或者当前num比栈顶元素小，就入栈；否则就说明栈顶元素右边第一个greater number就是当前元素，所以我们就将其插入哈希表中，同时栈顶元素出栈。循环这个过程直到栈为空或者当前num比栈顶元素小。最后别忘了将当前元素入栈。

最后再扫描一遍，对于findNums中的每个元素，在哈希表中查找它右侧的第一个greater number，并且加入结果集中。算法的时间复杂度是O(n)，空间复杂度是O(n)。

代码：

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        unordered_map<int, int> hash;
        vector<int> ret;
        stack<int> st;
        for (int i = 0; i < nums.size(); ++i) {
            if (st.empty() || st.top() > nums[i]) {
                st.push(nums[i]);
            }
            else {
                while (!st.empty() && st.top() < nums[i]) {
                    hash.insert(make_pair(st.top(), nums[i]));
                    st.pop();
                }
                st.push(nums[i]);
            }
        }
        for (int i = 0; i < findNums.size(); ++i) {
            if (hash.count(findNums[i]) == 0) {
                ret.push_back(-1);
            }
            else {
                ret.push_back(hash[findNums[i]]);
            }
        }
        return ret;
    }
};