标签:than nums Subsequence competitive stack Most subsequence needRemoveCount Find
package LeetCode_1673 import java.util.* /** * 1673. Find the Most Competitive Subsequence * https://leetcode.com/problems/find-the-most-competitive-subsequence/ * Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k. An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array. We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 is less than 5. Example 1: Input: nums = [3,5,2,6], k = 2 Output: [2,6] Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive. * */ class Solution { /* * solution: Stack,scan each number, check current num if large than then the first one in stack, * pop the large one, keep the increasing stack; * Time:O(n), Space:O(n) * */ fun mostCompetitive(nums: IntArray, k: Int): IntArray { var needRemoveCount = nums.size - k val stack = Stack<Int>() for (i in nums.indices) { while (stack.isNotEmpty() && nums[i] < stack.peek() && needRemoveCount > 0) { stack.pop() needRemoveCount-- } stack.push(nums[i]) } //handle case some case, for example 1111 while (needRemoveCount > 0) { stack.pop() needRemoveCount-- } val result = IntArray(k) var index = k - 1 //replace element from right to left,so no need to reverse while (stack.isNotEmpty()) { result[index--] = stack.pop() } return result } }
标签:than,nums,Subsequence,competitive,stack,Most,subsequence,needRemoveCount,Find 来源: https://www.cnblogs.com/johnnyzhao/p/14078118.html
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