排序数组中查找出现数字的个数
1:题目描述
统计一个数字在排序数组中出现的次数。
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: 2
示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: 0
限制:
0 <= 数组长度 <= 50000
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof
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2:题目分析
采用二分法查找定位,然后左右遍历统计个数。
3:代码示例
package JianZhiOffer53_1; import jdk.nashorn.internal.ir.ReturnNode; import javax.sound.midi.Soundbank; /** * @author :dazhu * @date :Created in 2020/4/6 16:40 * @description: * @modified By: * @version: $ */ public class Main { public static void main(String[]args){ Solution solution = new Solution(); int[] arr = new int[]{1,1,1,1,1,1,2,3,4,5,6,7,8,9,9,9,9,9,9}; System.out.println(solution.search(arr,9)); } } //先通过二分法找到定位,再左右扩展找到次数 class Solution { int junzhi = 0; int resultIndex = -1; public int search(int[] nums, int target) { return counterTarget(nums,binarySearch(nums,0,nums.length-1,target),target); } //迭代二分法,找到位置,找不到返回-1 //迭代终止条件:左指针大于等于右指针,或者已经找到target。 //迭代逻辑:取左右指针之平均,然后比较,如果大于target,则右指针赋值为平均,再递归 //如果小于target,则左指针赋值为平均,再递归。 public void binaryRecur(int[] nums, int left,int right,int target) { if(left<=right){ junzhi = (left+right)/2; if(nums[junzhi] > target ){ binaryRecur(nums, left,junzhi-1,target); } else{ if(nums[junzhi] < target ){ binaryRecur(nums, junzhi+1,right,target); } else{ resultIndex = junzhi; } } } } //不用迭代,用遍历来二分查找 public int binarySearch(int[] nums, int left,int right,int target) { int junzhi = 0; while(left<=right){ junzhi = (left+right)/2; if(nums[junzhi]>target){ right = junzhi-1; } else{ if(nums[junzhi]<target){ left = junzhi+1; } else{ return junzhi; } } } return -1; } //在所在定位,左右寻找,然后获得长度 public int counterTarget(int[] nums, int resultIndex,int target) { if(resultIndex == -1){ return 0; } else{ int tempIndex = resultIndex; int counter = 1; //向右寻找,要不能大于等于右边界 while(true){ if(resultIndex>=nums.length-1){ break; } resultIndex++; if(nums[resultIndex] != target){ break; } counter++; } //向左寻找,不能小于等于0 resultIndex =tempIndex; while(true){ if(resultIndex<=0){ break; } resultIndex--; if(nums[resultIndex] != target){ break; } counter++; } return counter; } } }
注意再二分法中遍历和递归条件都为left<=right,同时当进入新的一次查找时,要进行+1或者-1的操作。