Search for a Range

問題描述

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].

思考：如何得出target的左邊界和右邊界

想法

• 1、找出target第一次出現的index1和target+1第一次出現的index2，則{index1，index2-1}即爲所求；
• 2、找出A[mid] = target 的索引，然後二分法向左和右擴展

代碼

想法一：

public class Solution {
    public int[] searchRange(int[] A, int target) {    
        int[] ret = {-1,-1};  
        int left = find1thT(A,0,target);     
        if(left == A.length || A[left] != target)
            return ret;    
        ret[0] = left;
        ret[1] = find1thT(A,left + 1,target + 1) - 1;     
        return ret;
    }


    private int find1thT(int[] a, int star, int tar){       
        int start = star;     
        int end = a.length;   
        while(start < end){         
            int mid = start + ((end - start) >> 1);           
            if(tar > a[mid])
                start = mid + 1;
            else
                end = mid;  
        }  
        return start;
    }
}

想法二：

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int start = 0, end = A.length - 1;
        int left = -1, right = 0;
        while(start <= end){
            int mid = start + ((end - start) >> 1);
            if(target > A[mid])
                start = mid + 1;
            else if(target < A[mid])
                end = mid - 1;
            else{
                left = right = mid;
                //尋找左邊界
                while(start < left){
                    mid = start + ((left - start) >> 1);
                    if(target == A[mid]) left = mid;
                    else start = mid + 1;
                }
                //尋找右邊界
                while(right <= end){
                    mid = right + ((end - right) >> 1);
                    if(target == A[mid]) right = mid + 1;
                    else end = mid - 1;
                }
                break;
            }
        }
        int[] ret = {left,right - 1};
        return ret;
    }
}