34. Find First and Last Position of Element in Sorted Array

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Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

Follow up: Could you write an algorithm with O(log n) runtime complexity?

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

Constraints:

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums is a non-decreasing array.
• -109 <= target <= 109

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解法1

普通的BinarySearch 是找到相等的就返回结果. 但是本题中是要找到起始位置和终止位置.

所以, 分两步来做, 一次找到最左端, 一次找到最右端.

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = new int[]{-1, -1};
        
        if(nums.length == 0)
            return res;
        
        // 找到左边界
        int left = 0, right = nums.length - 1;
        while(left <= right){
            int mid = (left + right) / 2;
            // 目标值小于等于当前值时, 继续向左侧缩小范围
            if(target <= nums[mid])
                right = mid - 1;
            else
                left = mid + 1;
        }
        // 如果left超过边界,或者找到的值不等于目标值,说明目标值在数组中不存在
        if(left >= nums.length || nums[left] != target)
            return res;
        res[0] = left;
        
        // 找到右边界
        left = 0;
        right = nums.length - 1;
        while(left <= right){
            int mid = (left + right) / 2;
            if(target < nums[mid])
                right = mid - 1;
            // 目标值大于等于当前值时,继续向右侧缩小范围
            else
                left = mid + 1;
        }
        res[1] = right;
        
        return res;
    }
}