LeetCode 034 Search for a Range

题目描述

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

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

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

代码

    public static int[] searchRange(int[] nums, int target) {

        int l = 0, r = nums.length;

        while (l < r) {

            int m = l + (r - l) / 2;

            if (nums[m] == target) {

                int s = m, e = m;

                while (s - 1 >= 0 && nums[s - 1] == target) {
                    s--;
                }

                while (e + 1 < nums.length && nums[e + 1] == target) {
                    e++;
                }

                return new int[] { s, e };

            } else if (nums[m] > target) {
                r = m;
            } else {
                l = m + 1;
            }
        }

        return new int[] { -1, -1 };
    }
    原文作者:_我们的存在
    原文地址: https://blog.csdn.net/Yano_nankai/article/details/49755035
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