【LeetCode】496. Next Greater Element I

问题描述

问题链接：https://leetcode.com/problems/next-greater-element-i/#/description

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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.
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我的代码

看懂问题以后就很简单了，直接上代码。

public class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        /*
        思路是遍历nums1，对每一个数，找到它在nums2的位置，然后从这个位置开始向后遍历，
        如果能找到比它大的，就记下第一个比它大的数，否则记为-1
        */
        int len1 = nums1.length;
        int len2 = nums2.length;
        int[] result = new int[len1];

        for(int i = 0; i < len1; i++){
            int curNum = nums1[i];
            boolean startFlag = false;
            boolean findFlag = false;
            for(int j = 0; j < len2; j++){
                if(nums2[j] == curNum){
                    startFlag = true;
                    continue;
                }
                if(startFlag && nums2[j] > curNum){
                    result[i] = nums2[j];
                    findFlag = true;
                    break;
                }
            }
            if(!findFlag){
                result[i] = -1;
            }
        }
        return result;
    }
}

打败了10.69%的Java代码。

public class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        /*
        思路是遍历nums1，对每一个数，找到它在nums2的位置，然后从这个位置开始向后遍历，
        如果能找到比它大的，就记下第一个比它大的数，否则记为-1
        */
        int len1 = nums1.length;
        int len2 = nums2.length;
        int[] result = new int[len1];

        for(int i = 0; i < len1; i++){
            result[i] = -1;
            int curNum = nums1[i];
            boolean startFlag = false;

            for(int j = 0; j < len2; j++){
                if(nums2[j] == curNum){
                    startFlag = true;
                    continue;
                }
                if(startFlag && nums2[j] > curNum){
                    result[i] = nums2[j];
                    break;
                }
            }
        }
        return result;
    }
}

优化一下以后打败了15.76%。还是到讨论区看看老司机们吧。

讨论区

Java 10 lines linear time complexity O(n) with explanation

链接地址：https://discuss.leetcode.com/topic/77916/java-10-lines-linear-time-complexity-o-n-with-explanation

这个思路简直666。不说了，直接搬过来。

Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence

We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6

public int[] nextGreaterElement(int[] findNums, int[] nums) {
    Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
    Stack<Integer> stack = new Stack<>();
    for (int num : nums) {
        while (!stack.isEmpty() && stack.peek() < num)
            map.put(stack.pop(), num);
        stack.push(num);
    }   
    for (int i = 0; i < findNums.length; i++)
        findNums[i] = map.getOrDefault(findNums[i], -1);
    return findNums;
}

Easy to understand O(MN) Java solution

链接地址：https://discuss.leetcode.com/topic/77904/easy-to-understand-o-mn-java-solution

这个代码也是有点意思的，可以看看。

public class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        if(findNums == null ||  nums == null || 
           findNums.length == 0 || nums.length == 0 || 
           findNums.length > nums.length) return new int[0];

        int m = findNums.length;
        int n = nums.length;
        int[] result = new int[m];
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();

        for(int j = 0; j < n; ++j){
            map.put(nums[j], j);
        }
        for(int i = 0; i < m; ++i){
            int j = map.get(findNums[i]);
            for(; j < n; ++j){
                if(nums[j] > findNums[i]) break;
            }
            result[i] = j < n ? nums[j] : -1;
        }
        return result;
    }
}