[leetcode medium] 34. Find the first and last positions of elements in the sorted array - array double pointer

34. Find the first and last positions of the elements in the sort array

Pay attention to the handling of special situations

Method 1 ： Traverse

class Solution {
    
public:
    vector<int> searchRange(vector<int>& nums, int target) {
    
        vector<int> result = {
    -1,-1};
        // nums It's empty 
        if(nums.empty()) return result;
        int low = 0;
        int high = nums.size() - 1;
        // low < high : nums = [2, 2] target = 3
        while(nums[low] < target && low < high){
    
            low++;
        } 
        if(nums[low] != target) return result;
        while(nums[high] > target){
    
            high--;
        }
        result[0] = low;
        result[1] = high;
        return result;
    }
};

Method 2 ： Two points search

Because the whole array is non decreasing , You can use dichotomy to find The first is equal to （ Or greater than ） target The location of leftIdx and The first one is greater than target Minus one rightIdx

because nums May not exist in the array target , So we get two subscripts leftIdx and rightIdx check , See if they meet the requirements , If the conditions are met, return [leftIdx, rightIdx], Otherwise return to [-1,-1].

class Solution {
     
public:
	//  Two points search , if  lower  by  true, Find the first one greater than or equal to  target  The subscript , by  false  Then find the first one greater than  target  The subscript 
    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;
    }
	
    vector<int> searchRange(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 vector<int>{
    leftIdx, rightIdx};
        } 
        return vector<int>{
    -1, -1};
    }
};