本文介绍了排序和使用Java二进制搜索的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我被要求进行排序和搜索的数组。排序数组很简单,我的code的工作,但随后每当我试图把它适用于数组中的第一个元素,但给了我-1,结果
二进制搜索方法我的全code是如下:
公共静态无效的主要(字串[] args){ INT []数组=新INT [5]; 数组[0] = 50; 阵列[1] = 40; 阵列[2] = 10; 阵列[3] = 20; 数组[4] = 100;排序(数组,(array.length - 1)); 为(中间体X = 0; X&下; array.length; X ++){ 的System.out.println(+阵列[X]); } 的System.out.println(); 的System.out.println(二进制搜索(R):+ rBsearch(数组,0,(array.length),20));} 公共静态无效的排序(INT []一,INT最后){ 如果(最后一个大于0){ INT最大= findMax(一个,最后一个); 掉期(A,最后,最大值); 排序(A,最后 - 1); }}公共静态INT rBsearch(INT [] L,INT低,诠释高,诠释K){ INT中旬=(低+高)/ 2; 如果(低>的高点){ 返回-1; }否则如果(L [MID] == K){ 返回中旬; }否则如果(L [MID]< K){ 返回rBsearch(L,K,中+ 1,高); }其他{ 返回rBsearch(L,K,低,中 - 1); } }公共静态INT findMax(INT []编曲,诠释即止){ INT最大= 0; 的for(int i = 0; I< =最后,我++){ 如果(ARR [I]>常用3 [MAX]){ 最大= I; } } 返回最大值; }公共静态无效掉期(INT []改编,INT最后,诠释最大值){ INT TEMP = ARR [最后] ARR [最后] = ARR [MAX]; ARR [MAX] =温度;}解决方案
您疯玩了二进制搜索间隔
公共静态INT rBsearch(INT [] L,INT低,诠释高,诠释K){ INT中旬=(低+高)/ 2; 如果(低>的高点){ 返回-1; }否则如果(L [MID] == K){ 返回L [MID]; }否则如果(L [MID]< K){ 返回rBsearch(L,中+ 1,高,K); }其他{ 返回rBsearch(L,低,中 - 1,K); } }I was asked to sort and search an array. The sorting the array was simple and my code worked but then whenever I try to call the binary search method it works for the first element in the array but gives me "-1" as a result
My full code is as follows:
public static void main(String[] args) { int[] array = new int[5]; array[0] = 50; array[1] = 40; array[2] = 10; array[3] = 20; array[4] = 100; sort(array, (array.length - 1)); for (int x = 0; x < array.length; x++) { System.out.println(" " + array[x]); } System.out.println(""); System.out.println("Binary search (R): " + rBsearch(array, 0, (array.length), 20)); } public static void sort(int[] a, int last) { if (last > 0) { int max = findMax(a, last); swap(a, last, max); sort(a, last - 1); } } public static int rBsearch(int[] L, int low, int high, int k) { int mid = (low + high) / 2; if (low > high) { return -1; } else if (L[mid] == k) { return mid; } else if (L[mid] < k) { return rBsearch(L, k, mid + 1, high); } else { return rBsearch(L, k, low, mid - 1); } } public static int findMax(int[] arr, int last) { int max = 0; for (int i = 0; i <= last; i++) { if (arr[i] > arr[max]) { max = i; } } return max; } public static void swap(int[] arr, int last, int max) { int temp = arr[last]; arr[last] = arr[max]; arr[max] = temp; }解决方案
You goofed up the binary search intervals
public static int rBsearch(int[] L, int low, int high, int k) { int mid = (low + high) / 2; if (low > high) { return -1; } else if (L[mid] == k) { return L[mid]; } else if (L[mid] < k) { return rBsearch(L, mid + 1, high, k); } else { return rBsearch(L, low, mid - 1, k); } }
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排序和使用Java二进制搜索
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