Leetcode[java] 34. Search for a Range

最新推荐文章于 2021-03-26 19:54:40 发布

原创最新推荐文章于 2021-03-26 19:54:40 发布 · 175 阅读

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CC 4.0 BY-SA版权

本文介绍了一种算法，可在已排序的整数数组中查找给定目标值的起始和结束位置，采用二分查找思想实现O(log n)的时间复杂度。若未找到目标值，则返回[-1, -1]。

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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].

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = new int[]{-1,-1};
        int len = nums.length;
        if(len==0||nums==null)
            return res;
        if(target<nums[0] || target > nums[len-1])
            return new int[]{-1,-1};
        
        helper(target,nums,0,len-1,res);
        return res;
    }
    private void helper(int target, int[] nums, int begin, int end, int[] res){
        if(begin>end)
            return;

        if(nums[begin]==target&&nums[end]==target){
            res[0]=begin;
            res[1]=end;
            return;
        }
        int mid = begin + (end-begin)/2;
        if(target>nums[mid]){
             helper(target,nums,mid+1,end,res);
        } else if(target<nums[mid]){
            helper(target,nums,begin,mid-1,res);
        } else{
            // 向左右找
            res[0]=mid;
            res[1]=mid;
            int t = mid;
            while(t>begin && nums[t]==nums[t-1] ){
                t--;
                res[0]=t;
            }
            int t2 = mid;
            while(t2<end && nums[t2]==nums[t2+1]){
                t2++;
                res[1]=t2;
            }
            return;
        }
            
    }
}