【LeetCode】#34在排序数组中查找元素的第一个和最后一个位置(Find First and Last Position of Element in Sorted Array)

【LeetCode】#34在排序数组中查找元素的第一个和最后一个位置(Find First and Last Position of Element in Sorted Array)

题目描述

给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
你的算法时间复杂度必须是 O(log n) 级别。
如果数组中不存在目标值，返回 [-1, -1]。

示例

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]

Description

Given an array of integers nums sorted in ascending order, 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].

Example

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]

解法

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums.length==0){
            return new int[]{-1,-1};
        }
        int len=nums.length, l=0, m=len/2, r=len-1;
        int num1 = -1;
        while(true){
            int left=nums[l], mid=nums[m], right=nums[r];
            if(left==target){
                num1 = l;
                break;
            }
            if(mid==target){
                num1 = m;
                break;
            }
            if(right==target){
                num1 = r;
                break;
            }
            if(Math.abs(m-r)<=1 && Math.abs(m-l)<=1){
                return new int[]{-1,-1};
            }
            
            if(left<mid){
                if(left<target && target<mid){
                    r = m;
                    m = (l+r)/2;
                    continue;
                }else{
                    l = m;
                    m = (l+r)/2;
                    continue;
                }
            }
            
            if(mid<right){
                if(mid<target && target<right){
                    l = m;
                    m = (l+r)/2;
                    continue;
                }else{
                    r = m;
                    m = (l+r)/2;
                    continue;
                }
            }
        }
        int num2 = num1;
        while(true){
            if(nums[num1]==target){
                num1--;
            }else{
                num1++;
                break;
            }
            if(num1<0){
                num1 = 0;
                break;
            }
        }
        while(true){
            if(nums[num2]==target){
                num2++;
            }else{
                num2--;
                break;
            }
            if(num2>len-1){
                num2 = len-1;
                break;
            }
        }
        return new int[]{num1, num2};
    }
}