leetcode--面试题53 - I. 在排序数组中查找数字 I

面试题53 - I. 在排序数组中查找数字 I
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

注意：本题与主站 34 题相同（仅返回值不同）：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/

测试用例：
[]
1
[1]
1
[1]
0
[5,7,7,8,8,10]
8
[5,7,7,8,8,10]
6

思路：二分法

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size(), start = 0, end = 0;
        int left = 0, right = n, mid = 0;
        while (left < right) {
            mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                right = mid;
            } else if (nums[mid] > target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        if (left == n) return 0;
        if (nums[left] == target) start = left;
        else return 0;
        left = 0, right = n, mid = 0;
        while (left < right) {
            mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        //if (left == 0) return 0;
        if (nums[left-1] == target) end = left - 1;
        //else return 0;
        return end - start + 1;
    }
};
/*12ms,15.7MB*/

比上面解法效果进一步优化：

class Solution {
    public int result = 0;
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        while (left >= 0 && right <= nums.length - 1 && left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                result = 1;
                int _left = mid - 1;
                int _right = mid + 1;
                while (_left >= left) {
                    if (nums[_left--] == target) result++;
                    else break;
                }
                while (_right <= right) {
                    if (nums[_right++] == target) result++;
                    else break;
                }
                break;
            }
        }
        return result;
    }
}

作者：le-ge-zhi-shang
链接：https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/solution/shuang-100ji-bai-er-fen-cha-zhao-fa-by-le-ge-zhi-s/
来源：力扣（LeetCode）
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