刷题记录——剑指 Offer 53 - I. 在排序数组中查找数字 I

2021/9/27

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

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

二分查找。当nums[mid]等于target时，说明数组中存在该数字，再从mid的位置向两边走，计算该数字出现的起始位置begin和终止位置end，位置下标之差加一就是该数字出现的次数。

class Solution {
public:
    int search(vector& nums, int target) {
        if(nums.size() < 0 || nums.size()>100000)
            return -1;
        int left = 0;
        int right = nums.size() - 1;
        int mid = (right-left)/2 +left;
        int pos = mid;
        while(left <= right){
            if(nums[mid] == target){
                pos = mid;                
                while(mid>=0 && nums[mid] == target) {
                    left = mid;
                    mid--;
                }   
                mid = pos; 
                while(mid < nums.size() && nums[mid] == target)  {
                    right = mid;
                    mid++;
                }
                break;
            }else if(nums[mid] > target){
                right = mid -1;                
            }else{
                left = mid + 1;
            }
            mid = (right-left)/2 +left;
        }
        return (right-left+1);
    }
};

官方题解给出的题解：直接用二分法找边界。分别查找target的右边界和target-1 的右边界，两个边界值之差就是target出现的次数。（代码可以复用了）
算法流程：
1.初始化二分查找的左边界left=0,右边界right=nums.length-1.
mid = (right-left)/2 +left
2.若nums[mid]>target，右边界位于[left, mid-1]，令right=mid-1；若nums[mid]<=target，右边界位于[mid+1 right]，令left= mid+1.

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