Leetcode Search for a Range

最新推荐文章于 2018-09-05 14:58:19 发布

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最新推荐文章于 2018-09-05 14:58:19 发布

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分类专栏： Leetcode Leetcode 重启

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本文介绍了一个算法，用于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，采用二分搜索实现，确保了算法的时间复杂度为O(log n)。

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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].

Difficulty:Medium

public class Solution {
    public int binSearch(int[] nums, int start, int end, int target){
        if(start > end || target > nums[end] || target < nums[start])
            return -1;
        int mid = start + (end - start)/2;
        if(nums[mid] == target)
            return mid;
        if(nums[start] == target)
            return start;
        if(nums[end] == target)
            return end;
        if(nums[mid] < target && target < nums[end]){
            return binSearch(nums, mid + 1, end - 1, target);
        }
        else{
            return binSearch(nums, start + 1, mid - 1, target);
        }
    }
    public int[] searchRange(int[] nums, int target) {
        int ans = binSearch(nums, 0, nums.length - 1, target);
        int data[] = new int[2];
        data[0] = -1; data[1] = -1;
        if(ans == -1)
            return data;
        int start = ans;
        int end = ans;
        while(start >= 0 && nums[start] == target ){
            start--;
        }
        start++;
        while(end < nums.length && nums[end] == target ){
            end++;
        }
        end--;
        data[0] = start; data[1] = end;
        return data;
    }
}