剑指 Offer 53 - I. 在排序数组中查找数字 I

剑指 Offer 53 - I. 在排序数组中查找数字 I

题目

链接

https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/

问题描述

统计一个数字在排序数组中出现的次数。

0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums 是一个非递减数组
-109 <= target <= 109

示例

输入: nums = [5,7,7,8,8,10], target = 8
输出: 2

提示

思路

递增数组，找到某个数的出现次数，那么用二分查找找到左右边界，直接计算即可。

复杂度分析

时间复杂度 O(logn)
空间复杂度 O(1)

代码

Java

    public static int search(int[] nums, int target) {
        if (nums.length == 0) {
            return 0;
        }
        int left = 0, right = nums.length - 1;
        //左边界
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] > target){
                right = mid - 1;
            }else if(nums[mid] < target){
                left = mid + 1;
            }else if(nums[mid] == target){
                right = mid - 1;
            }
        }
        int ans = left;
        //右边界
        left = 0; right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] > target){
                right = mid - 1;
            }else if(nums[mid] < target){
                left = mid + 1;
            }else if(nums[mid] == target){
                left = mid + 1;
            }
        }
        ans = right - ans + 1;
        return ans;
    }