[C++知识库]剑指 Offer 53 - I. 在排序数组中查找数字 I C++ 2021-08-16

剑指 Offer 53 - I. 在排序数组中查找数字 I

1、直观感受，二分找到Target位置，然后向两边扩展计数。

class Solution {
public:
    int numPosition(vector<int>& nums, int target)
    {
        int left = 0;
        int right = nums.size();
        int mid = left + (right - left)/2;

        while (left < right)
        {
            if (nums[mid] == target)
                return mid;
            
            if  (target > nums[mid])
            {
                left = mid + 1;
            }
            else
            {
                right = mid;
            }
            mid = left + (right - left)/2;
        }
        return -1;
    }

    int search(vector<int>& nums, int target) 
    {
        int pos = numPosition(nums, target);
        
        if (pos < 0)
        {
            return 0;
        }

        int temp = pos;
        int count = 0;
        while (temp >= 0 && nums[temp--] == target)
            count++;

        temp = pos + 1;
        while (temp < nums.size() && nums[temp++] == target)
            count++;
         
        return count;
    }
};

2、二分优化，直接找两端

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

    int search(vector<int>& nums, int target) 
    {
        int leftIdx = binarySearch(nums, target, true);
        int rightIdx = binarySearch(nums, target, false) - 1;
        if (leftIdx <= rightIdx && rightIdx < nums.size() && nums[leftIdx] == target && nums[rightIdx] == target) 
        {
            return rightIdx - leftIdx + 1;
        }
        return 0;
    }
};