LeetCode 440. K-th Smallest in Lexicographical Order (Java版; Hard)

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LeetCode 440. K-th Smallest in Lexicographical Order (Java版; Hard)

题目描述

Given integers n and k, find the lexicographically k-th smallest integer in the range from 1 to n.

Note: 1 ≤ k ≤ n ≤ 109.

Example:

Input:
n: 13   k: 2

Output:
10

Explanation:
The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.

第一次做; 直接看的leetcode题解; 核心:字典序排列也就是对十叉树进行先序遍历

class Solution {
    public int findKthNumber(int n, int k) {
        //根据题意, 第一个数是1
        int cur = 1;
        //第一个是1, 所以再找出k-1个数后就知道第k个数是多少了
        k = k-1;
        while(k>0){
            int steps = calSteps(n, cur, cur+1);
            //横向扩展, 相当于+steps,
            if(steps<=k){
                cur +=1 ;
                k -= steps;
            }
            //steps > k; 纵向扩展, 相当于+1
            else{
                cur *= 10;
                k-=1;
            }
            
        }
        return cur;
    }
    
    //n1 和 n2得是long类型的, int会产生溢出, 不能通过这个案例: 输入n=681692778, k=351251360, 预期结果=416126219
    private int calSteps(int n, long n1, long n2){
        int steps = 0;
        while(n1<=n){
            steps += Math.min(n2, n+1) - n1;
            n1 *= 10;
            n2 *= 10;
        }
        return steps;
    }
}

第一次做; 超时, 29/69

/*
小根堆
*/
class Solution {
    public int findKthNumber(int n, int k) {
        //
        PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>(new Comparator<Integer>(){
            public int compare(Integer o1, Integer o2){
                String str1 = String.valueOf(o1);
                String str2 = String.valueOf(o2);
                return str1.compareTo(str2);
            }
        });
        for(int i=1; i<=n; i++){
            minHeap.add(i);
        }
        for(int i=0; i<k-1; i++){
            minHeap.poll();
        }
        return minHeap.peek();
    }
}

leetcode最优解

public int findKthNumber(int n, int k) {
    int curr = 1;
    k = k - 1;
    while (k > 0) {
        int steps = calSteps(n, curr, curr + 1);
        if (steps <= k) {
            curr += 1;
            k -= steps;
        } else {
            curr *= 10;
            k -= 1;
        }
    }
    return curr;
}
//use long in case of overflow
public int calSteps(int n, long n1, long n2) {
    int steps = 0;
    while (n1 <= n) {
        steps += Math.min(n + 1, n2) - n1;
        n1 *= 10;
        n2 *= 10;
    }
    return steps;
}