Find the Kth Largest Integer in the Array

最新推荐文章于 2022-10-27 17:29:32 发布

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最新推荐文章于 2022-10-27 17:29:32 发布

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分类专栏： quick select 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/122613735

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本文介绍了一种使用快速选择算法找出字符串数组中第K大的整数的方法。该算法适用于处理包含重复数字的情况，并通过比较字符串长度及字典序实现了非递减排序。文章详细解释了算法流程并提供了实现代码。

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You are given an array of strings nums and an integer k. Each string in nums represents an integer without leading zeros.

Return the string that represents the kth largest integer in nums.

Note: Duplicate numbers should be counted distinctly. For example, if nums is ["1","2","2"], "2" is the first largest integer, "2" is the second-largest integer, and "1" is the third-largest integer.

Example 1:

Input: nums = ["3","6","7","10"], k = 4
Output: "3"
Explanation:
The numbers in nums sorted in non-decreasing order are ["3","6","7","10"].
The 4th largest integer in nums is "3".

Example 2:

Input: nums = ["2","21","12","1"], k = 3
Output: "2"
Explanation:
The numbers in nums sorted in non-decreasing order are ["1","2","12","21"].
The 3rd largest integer in nums is "2".

Example 3:

Input: nums = ["0","0"], k = 2
Output: "0"
Explanation:
The numbers in nums sorted in non-decreasing order are ["0","0"].
The 2nd largest integer in nums is "0".

Constraints:

1 <= k <= nums.length <= 104
1 <= nums[i].length <= 100
nums[i] consists of only digits.
nums[i] will not have any leading zeros.

思路: quick select. O(n*M); M is word length;

class Solution {
    public String kthLargestNumber(String[] nums, int k) {
        return String.valueOf(findKth(nums, 0, nums.length - 1, k));
    }
    
    private String findKth(String[] A, int start, int end, int k) {
        if(start == end) {
            return A[start];
        }
        int i = start; int j = end;
        int mid = start + (end - start) / 2;
        String pivot = A[mid];
        while(i <= j) {
            while(i <= j && compare(A[i], pivot) > 0) {
                i++;
            }
            while(i <= j && compare(A[j], pivot) < 0) {
                j--;
            }
            if(i <= j) {
                String temp = A[i];
                A[i] = A[j];
                A[j] = temp;
                i++;
                j--;
            }
        }
        
        if(start + k - 1 <= j) {
            return findKth(A, start, j, k);
        }
        if(start + k - 1 >= i) {
            return findKth(A, i, end, k - (i - start));
        }
        return A[j + 1];
    }
    
    private int compare(String a, String b) {
        if(a.length() == b.length()) {
            return a.compareTo(b);
        } else {
            return a.length() - b.length();
        }
    }
}