maximum-gap(经过了提示)
下面的分桶个数做的不太好,原来的解法是用的
int gap = (big - small) / vlen;
if (gap == 0) {
gap = 1;
}
下面是现在的Java解法:
package com.company;
import java.util.*;
class Solution {
public int maximumGap(int[] nums) {
if (nums.length < 2) {
return 0;
}
// 用 Radix 排序
int small = Integer.MAX_VALUE;
int big = 0;
for (int num : nums) {
if (num < small) {
small = num;
}
if (num > big) {
big = num;
}
}
System.out.println("big is " + big + " small " + small);
int gap = (big - small - 1) / (nums.length - 1) + 1;
if (gap == 0) {
return 0;
}
int gap_num = (big - small) / gap + 1;
int[] first = new int[gap_num];
int[] second = new int[gap_num];
// [ )
System.out.println("gap is " + gap + " len is " + nums.length + "big is " + big + " small " + small);
for (int num : nums) {
int index = (num - small) / gap;
if (first[index] == 0 || num < first[index]) {
first[index] = num;
}
if (second[index] == 0 || num > second[index]) {
second[index] = num;
}
}
int ret = -1;
int last = -1;
for (int i=0; i<gap_num; i++) {
if (first[i] == 0) {
continue;
}
if (last == -1) {
last = second[i];
ret = last - first[i];
}
else {
if (first[i] - last > ret) {
ret = first[i] - last;
}
if (second[i] - first[i] > ret) {
ret = second[i] - first[i];
}
last = second[i];
}
}
return ret;
}
}
public class Main {
public static void main(String[] args) {
System.out.println("Hello!");
Solution solution = new Solution();
int[] nums = {1,1,1,1,1,5,5,5,5,5};
int ret = solution.maximumGap(nums);
System.out.printf("Get ret: %s\n", ret);
/*Iterator<List<Integer>> iterator = ret.iterator();
while (iterator.hasNext()) {
Iterator iter = iterator.next().iterator();
while (iter.hasNext()) {
System.out.printf("%d,", iter.next());
}
System.out.println();
}*/
System.out.println();
}
}
maximum-gap(经过了提示)的更多相关文章
- 【leetcode】Maximum Gap
Maximum Gap Given an unsorted array, find the maximum difference between the successive elements in ...
- [LintCode] Maximum Gap 求最大间距
Given an unsorted array, find the maximum difference between the successive elements in its sorted f ...
- 线性时间的排序算法--桶排序(以leetcode164. Maximum Gap为例讲解)
前言 在比较排序的算法中,快速排序的性能最佳,时间复杂度是O(N*logN).因此,在使用比较排序时,时间复杂度的下限就是O(N*logN).而桶排序的时间复杂度是O(N+C),因为它的实现并不是基于 ...
- Maximum Gap
Given an unsorted array, find the maximum difference between the successive elements in its sorted f ...
- leetcode[164] Maximum Gap
梅西刚梅开二度,我也记一题. 在一个没排序的数组里,找出排序后的相邻数字的最大差值. 要求用线性时间和空间. 如果用nlgn的话,直接排序然后判断就可以了.so easy class Solution ...
- 【leetcode 桶排序】Maximum Gap
1.题目 Given an unsorted array, find the maximum difference between the successive elements in its sor ...
- 【LeetCode】164. Maximum Gap (2 solutions)
Maximum Gap Given an unsorted array, find the maximum difference between the successive elements in ...
- 由Maximum Gap,对话桶排序,基数排序和统计排序
一些非比较排序 在LeetCode中有个题目叫Maximum Gap.是求一个非排序的正数数列中按顺序排列后的最大间隔.这个题用桶排序和基数排序都能够实现.以下说一下桶排序.基数排序和计数排序这三种非 ...
- LeetCode 164. Maximum Gap[翻译]
164. Maximum Gap 164. 最大间隔 Given an unsorted array, find the maximum difference between the successi ...
- 【刷题-LeetCode】164 Maximum Gap
Maximum Gap Given an unsorted array, find the maximum difference between the successive elements in ...
随机推荐
- mysql存储过程出现OUT or INOUT argument 10 for routine
OUT or INOUT argument 10 for routine * is not a variable or NEW pseudo-variable 我查网上很多出现在call的时候没有添加 ...
- android开发 更新升级安装到一半自动闪退
如题:android开发 更新升级安装到一半自动闪退,,,解决办法,如下(红色为我新增的代码) /** * 安装APK文件 */ private void installApk( ...
- 下拉刷新ListView实现原理
(1)主要是onScroll()方法和onTouchEvent()方法,先是onTouchEvent()的ACTION_DOWN,然后是 ACTION_MOVE和onScroll()方法同时进行,最后 ...
- 1502: [NOI2005]月下柠檬树 - BZOJ
Description Input 文件的第1行包含一个整数n和一个实数alpha,表示柠檬树的层数和月亮的光线与地面夹角(单位为弧度).第2行包含n+1个实数h0,h1,h2,…,hn,表示树离地的 ...
- python 日期转星期
import time import datetime today = int(time.strftime('%w')) print today anyday = datetime.datetime( ...
- ios 开发常用快捷键
CTRL + K 删除一行,尽量在行首处使用: CMD+ / 注释,取消注释 CMD + R 运行 CMD + . 停止运行 CMD + F 普通搜索 CMD + CTRL + ↑/↓ 切换头 ...
- 输入一个递增排序的数组和一个数字S,在数组中查找两个数,使得他们的和正好是S,如果有多对数字的和等于S,输出两个数的乘积最小的。
// test20.cpp : 定义控制台应用程序的入口点. // #include "stdafx.h" #include<iostream> #include< ...
- 剑指offer--面试题14
#include "stdafx.h" #include <iostream> using namespace std; //调整数组顺序使奇数位于偶数前 void O ...
- Ioc 比较
public interface IC { } public class A { IC ic_; public A(IC ic) { ic_ = ic; } } public class B : IC ...
- linux “命令行自动补全”功能用命令
是按Tab键,左上角ESC的下面两个,如果你当前目录只有一项,只需要直接Tab,如果有多项,输入前面不同的部分再Tab,一般输入3个字母就可以,如果按一下没效果,按两下会列出所有项,然后再输入一点自己 ...