Leetcode-34. Search for a Range

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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].

这个题目一开始没注意审题，即如果数组中没有这个数就返回[-1,1]，因此我们需要做的第一步是先找有没有这个数。第二步和第三步我的做法是由于都是整数，其实变成找target-0.5和target+0.5就可以了。Your runtime beats 10.26% of java submissions.

public class Solution {    public int[] searchRange(int[] nums, int target) {        if(nums[0] > target || nums[nums.length - 1] < target) return new int[]{-1,-1};if(nums.length == 1 ) return new int[]{0,0};if(nums[0] == target && nums[nums.length-1] == target) return new int[]{0,nums.length-1};int l_position = 0, r_position = 0;int l = 0, r = nums.length - 1;while(r - l > 1){int mid = (l + r)/2;if(nums[mid]<target) l = mid;else if(nums[mid] > target) r = mid;else break;}if(nums[(l+r)/2] != target && nums[l] != target && nums[r] !=target) return new int[]{-1,-1};if(nums[0] == target) l_position = 0;else{while( r - l > 1){int mid = (l + r) / 2;if(nums[mid] < target - 0.5) l = mid;elser = mid;}l_position = l + 1;}r = nums.length - 1;if(nums[nums.length - 1] == target ) r_position = nums.length - 1;else{while( r- l > 1){int mid = (l + r )/2;if(nums[mid] < target + 0.5) l = mid;else r = mid;}r_position = r - 1;}return new int[]{l_position, r_position};    }}

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