【重点】【二分查找】34. 在排序数组中查找元素的第一个和最后一个位置

题目
剑指中重题

Python

时间复杂度：O(logn)，其中 n 为 nums 的长度。
空间复杂度：O(1)，仅用到若干额外变量。

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        start = self.find(nums, target)
        end = self.find(nums, target + 1) - 1

        return [start, end] if start <= end else [-1, -1]
    
    def find(self, nums, target):
        l, r = 0, len(nums)-1
        while l <= r: # 要考虑mid=l(r)的情况，所以必须保留等号
            mid = (l + r) // 2
            if nums[mid] < target: 
                l = mid + 1 # 肯定满足 nums[l] >= target
            else:
                r = mid - 1
        
        return l

Java

法1：二分查找

注意：寻找完边界后，要判断两者是否合法！！！

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums.length == 0) {
            return new int[]{-1, -1};
        }
        int leftBound = findLeftBound(nums, target, 0, nums.length - 1);
        int rightBound = findRightBound(nums, target, 0, nums.length - 1);
        if (leftBound <= rightBound) {
            return new int[]{leftBound, rightBound};
        }

        return new int[]{-1, -1};
    }

    public int findLeftBound(int[] nums, int target, int left, int right) {
        int mid = 0;
        while (left <= right) {
            mid = left + (right - left) / 2;
            if (nums[mid] >= target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return left;
    }

    public int findRightBound(int[] nums, int target, int left, int right) {
        int mid = 0;
        while (left <= right) {
            mid = left + (right - left) / 2;
            if (nums[mid] <= target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return right;
    }
}