34. Search for a Range

Posted 2020-08-09 Machelsky

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].

 

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思路：两个bianry search找左右边界。左边界if(nums[mid]==target)，继续搜索左边high=mid;右边界nums[mid]==target,继续搜索右边，low=mid.提交后运行结果是个死循环，22的情况就会一直死循环。参考了discussion的一种方法，mid=low+(high-low+1)/2，biase to right.这样结果就没问题了

ref:https://discuss.leetcode.com/topic/59880/one-solution-one-is-two-binary-search-the-other-worst-case-o-n

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int l1=0;
        int h1=nums.length-1;
        int[] res={-1,-1};
        while(l1<h1)
        {
            int mid=l1+(h1-l1)/2;
            if(nums[mid]>=target)
            {
                h1=mid;
            }
            else
            {
                l1=mid+1;
            }
        }
        int l2=0;
        int h2=nums.length-1;
        while(l2<h2)
        {
            int mid=l2+(h2-l2+1)/2;
            if(nums[mid]<=target)
            {
                l2=mid;
            }
            else
            {
                h2=mid-1;
            }
        }
        if(l1<=l2&&nums[l1]==target)
        {
            res[0]=l1;
            res[1]=l2;
        }
        return res;
    }
}