leetcode -- Search for a Range (TODO)

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].

思路：

使用二分搜索找到target的idx，然后查看该idx的左右确定范围。

算法复杂度：

平均情况下是O(lgn);

最坏情况下数组中所有元素都相同O(n);

public class Solution {
    public int[] searchRange(int[] A, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int idx = binarySearch(A, target);
        int len = A.length;
        int[] results = null;
        if(idx == -1){
            results = new int[]{-1, -1};
        } else{
            int l = idx;
            int r = idx;
            while(l >= 0 && A[l] == target){
                l--;
            }
            l++;
            
            while(r < len && A[r] == target){
                r++;
            }
            r--;
            results = new int[]{l, r};
        }
        return results;
    }
    
    public int binarySearch(int[] A, int target){
        int len = A.length;
        int l = 0, r = len - 1;
        while(l <= r){
            int mid = (l + r) / 2;
            if(target == A[mid])
                return mid;
            
            if(target > A[mid]){
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        
        return -1;
    }
}

 

google了下，要保证最坏情况下时间复杂度为O(lgn)：进行两次二分搜索确定左右边界