Algotithem_BinarySearch BinarySearch BinarySearch Given an array of integers nums which is sorted in...解法 一开始我的想法是:类似BinarySearch,先从中间值开始,如果中间的是 BadVersion,则继续往前取中间;如果中间的不是 BadVersion,则继续往后取中间。...解法 这个解法和 BinarySearch 逻辑一样,唯一不同的是,BinarySearch查找不到返回-1,而这个查找不到相等的最后返回的是 left 的位置。
You are given a list of integers nums representing coordinates of houses on a 1-...
binarySearch(Object[], Object key) 方法的object[]参数是要查找的数组,key参数为要查找的key值。...举例: int a[] = new int[] { 1, 3, 4, 6, 8, 9 }; int x1 = Arrays.binarySearch(a, 5);...int x2 = Arrays.binarySearch(a, 4); int x3 = Arrays.binarySearch(a, 0);...int x4 = Arrays.binarySearch(a, 10); System.out.println("x1:" + x1 + ", x2:" + x2);...); int x2 = Arrays.binarySearch(a, 1, 4, 4); int x3 = Arrays.binarySearch(a, 1, 4
二分搜索BinarySearch的 “来龙去脉” 二分搜索用于检索某个key是否在已排好序的序列中,我们还记得上编程语言的基础课程:猜字游戏吗?
参考链接: Java中的二分搜索binarySearch binarySearch()方法提供了多种重载形式,用于满足各种类型数组的查找需要,binarySearch()有两种参数类型 注:此法为二分搜索法...⑴.binarySearch(object[ ], object key); 如果key在数组中,则返回搜索值的索引;否则返回-1或者”-“(插入点)。...(a, 5); int x2 = Arrays.binarySearch(a, 4); int x3 = Arrays.binarySearch(a, 0); ...⑵.binarySearch(object[ ], int fromIndex, int endIndex, object key); 如果要搜索的元素key在指定的范围内,则返回搜索键的索引;否则返回...(a, 1, 4, 5); int x2 = Arrays.binarySearch(a, 1, 4, 4); int x3 = Arrays.binarySearch(
思维二分题1.在排序数组中查找元素第一个和最后一个位置2.寻找丑数3.思维题二分—Hamburgers
@Test public void binarySearch() { //数组一定要是顺序的。
package com.cn.search; import java.util.Scanner; public class BinarySearch { public void binarySearch...; } public static void main(String[] args) { BinarySearch binarySearch = new BinarySearch(); int...the search number:"); Scanner scanner = new Scanner(System.in); int search = scanner.nextInt(); binarySearch.binarySearch
static int binarySearch(long[] a, long key) static int binarySearch(Object[] a, int fromIndex, int toIndex..., Object key) static int binarySearch(Object[] a, Object key) static int binarySearch(short[] a, int...long key) { return binarySearch0(a, 0, a.length, key); } public static int binarySearch(long[] a...(a, fromIndex, toIndex, key); } 从代码中可以看到,两个方法最后都调用了 binarySearch0() 方法,但是在第二个binarySearch() 方法中调用了..., key, c); } 以上两个方法同样调用了 binarySearch0() 方法,该 binarySearch0() 方法的实现代码如下: private static int binarySearch0
但是java的Arrays.binarySearch()方法,如果集合中有重复元素,而且遇到目标元素正好是这些重复元素之一,该方法只能返回一个,并不能将所有的重复目标元素都返回,没办法,只能自造轮子了。...先复习下二分查找的经典算法: 1 private int binarySearch1(Integer[] A, Integer x) { 2 int low = 0, high...=> "); 4 System.out.println(binarySearch1(A, 5)); 5 6 System.out.println("binarySearch2...=> "); 7 System.out.println(binarySearch2(A, 5)); binarySearch1 => 5 binarySearch2 => [4,...=> "); 15 System.out.println(binarySearch4(awbList, x, comparator)); binarySearch4 => [3, 4
二分查找模板 def binarySearch(nums, target): left = 0 right = len(nums) - 1 while left <= right...(nums, 7) print(res) 寻找左侧边界的二分查找 def binarySearch(nums, target): left = 0 right = len(nums)...= target: return -1 return left nums = [1,1,2,2,2,3] res = binarySearch(nums, 2) print(res...) 寻找右侧边界的二分查找 def binarySearch(nums, target): left = 0 right = len(nums) while left < right...= target: return -1 return right - 1 nums = [1,2,2,2,2,3,3,4] res = binarySearch(nums, -
int[] array = { 8, 11, 21, 28, 32, 43, 48, 56, 69, 72, 80, 94 }; Console.WriteLine(BinarySearch...(array, 80)); Console.WriteLine(BinarySearch(array, 66, 0, array.Length - 1)); Console.ReadKey...(); } private static int BinarySearch(int[] array, int key) { //直接求解 var min...return mid; } } return -1; } private static int BinarySearch...(array, key, low, mid - 1); else if (array[mid] < key) return BinarySearch(array,
代码清单 1 def _binarySearch(key,a,lo,hi):if hikey: #中间位置项目大于查找关键字return_binarySearch(key,a,lo,mid) #递归查找前一子表elif a[mid]<key: #中间位置项目小于查找关键字...return_binarySearch(key,a,mid+1,hi) #递归查找后一子表else: #中间位置项目等于查找关键字return mid #查找成功,返回下标位置...def binarySearch(key,a) #二分查找return_binarySearch(key,a,0,len(a)) #递归二分查找法def main():a=[1,13,26,33,45,55,68,72,83,99...]print("关键字位于列表索引",binarySearch(33,a))#二分查找关键字33print("关键字位于列表索引",binarySearch(58,a))#二分查找关键字58if__name
search(int[] nums, int target) { // 注意第一次调用right为nums.length-1,而不是nums.length return BinarySearch...(nums, target, 0, nums.length-1); } public int BinarySearch(int [] nums,int target, int left...mid] == target) { return mid; } else if(nums[mid] < target) { return BinarySearch...(nums, target, mid + 1, right); } else if(nums[mid] > target) { return BinarySearch...(nums, target, 0, nums.length-1); } public int BinarySearch(int [] nums,int target, int left
(arr,1); int b = Arrays.binarySearch(arr,2); int c = Arrays.binarySearch(arr,3);...int d = Arrays.binarySearch(arr,4); int e = Arrays.binarySearch(arr,5); int f = Arrays.binarySearch...(arr,6); int g = Arrays.binarySearch(arr,7); int h = Arrays.binarySearch(arr,8);...(arr1,1); b = Arrays.binarySearch(arr1,2); c = Arrays.binarySearch(arr1,3); d...= Arrays.binarySearch(arr1,4); e = Arrays.binarySearch(arr1,5); f = Arrays.binarySearch
********/ 正常的解题思路: #include #include #include int arr[1000000]; int binarySearch...(arr,0,n-1)); } return 0; } int binarySearch(int *arr,int front,int rear){ if(front+1 ==...int find1 = binarySearch(arr,front,index); int find2 = binarySearch(arr,index+1,rear);...find1:find2; }else if(arr[index] >= arr[front] && arr[index] > arr[rear]) binarySearch(arr...,index,rear); else binarySearch(arr,front,index); } /************************************
public int binarySearch(int[] nums, int K) { int l = 0, h = nums.length - 1; while (l <= h) {...public int GetNumberOfK(int[] nums, int K) { int first = binarySearch(nums, K); int last = binarySearch...0 : last - first; } 需要注意以上实现的查找第一个位置的 binarySearch 方法,h 的初始值为 nums.length,而不是 nums.length - 1。...这是因为 binarySearch 只会返回 [0, nums.length - 1] 范围的值,对于 binarySearch([2,2], 3) ,我们希望返回 3 插入 nums 中的位置,也就是数组最后一个位置再往后一个位置...所以我们需要将 h 取值为 nums.length,从而使得 binarySearch 返回的区间更大,能够覆盖 k 大于 nums 最后一个元素的情况。
function binarySearch(arr, target) { let left = 0; // 数组第一个位置 let right = arr.length -1; // 数组中最后一个位置...(v => v === target) } binarySearch([1,3,4,5,7,8], 3); // 1 你会发现原生提供的findIndex无论数组中是否有序,还是无序都可以找到target...暴力for循环找索引 function binarySearch(arr = [], target) { let index = target ?...([1,3,4,5,7,8], 3); // 1 巧用map,移花接木 map这种方式的缺陷是数组中不能有重复的值,只是针对无重复的数组 function binarySearch(arr = [],...map.get(target) : -1 } binarySearch([1,3,4,5,7,8], 3); // 1 借用对象 只针对无重复数组 function binarySearch(arr
int search(vector& nums, int target) { if (nums.empty()) return -1; return binarySearch...(0, (nums.size() - 1) / 2, nums.size() - 1, nums, target); } int binarySearch(int begin, int...(begin, (begin + mid - 1) / 2, mid - 1, nums, target)>=0) { return binarySearch(begin..., (begin + mid - 1) / 2, mid - 1, nums, target); } else if(nums[mid]=0) { return binarySearch(mid +
二分查找 private int[] array = {1,2,2,3,4,6}; private int min = -20; private int max = 8; public void binarySearch...() { int left = binarySearch(array,min,true); int right = binarySearch(array,max,false);...System.out.println(left +" "+ right); } private int binarySearch(int[] array,int target,boolean first
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