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  • 503. Next Greater Element II

    package LeetCode_503
    
    import java.util.*
    
    /**
     * 503. Next Greater Element II
     * https://leetcode.com/problems/next-greater-element-ii/description/
     *
     * Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element.
     * The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number.
     * If it doesn't exist, output -1 for this number.
    
    Example 1:
    Input: [1,2,1]
    Output: [2,-1,2]
    Explanation: The first 1's next greater number is 2;
    The number 2 can't find next greater number;
    The second 1's next greater number needs to search circularly, which is also 2.
    
    Note: The length of given array won't exceed 10000.
     * */
    class Solution {
        /**
         * solution 1: brute force, Time complexity:O(n*n), Space complexity:O(n)
         * solution 2: monotonic stack, Time complexity:O(n), Space complexity:O(n)
         * */
        fun nextGreaterElements(nums: IntArray): IntArray {
            val n = nums.size
            val result = IntArray(n, { -1 })
            //solution 1
            /*for (i in nums.indices) {
                //because need loop through array, we use j%n
                for (j in i + 1 until (i + n)) {
                    if (nums[j % n] > nums[i]) {
                        result.set(i, nums[j % n])
                        break
                    }
                }
            }*/
            //solution 2, store indices of monotonically increasing element
            val stack = Stack<Int>()
            for (i in 0 until 2 * n) {
                //because need loop through array, we use j%n
                val num = nums[i % n]
                while (stack.isNotEmpty() && nums[stack.peek()] < num) {
                    result.set(stack.pop(),num)
                }
                //need care about the index
                if (i<n) {
                    stack.push(i)
                }
            }
            return result
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13124408.html
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