【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-CSDN博客

本文链接：https://blog.csdn.net/Allenlzcoder/article/details/135091456

本文介绍了使用Java编写的二分查找算法，解决在给定整数数组中找到目标值的左右边界问题，如果目标值存在则返回边界，否则返回-1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Python

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        n = len(nums)
        start = self.lower_bound(nums, 0, n-1, target)
        end = self.lower_bound(nums, 0, n-1, target + 1) - 1

        return [-1, -1] if start > end else [start, end]

    def lower_bound(self, nums, l, r, target): # 寻找 >= target的第一个数!!!
        while l <= r:
            mid = (l + r) // 2
            if nums[mid] < target:
                l = mid + 1
            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;
    }
}