[LintCode] 61 Search for a Range 解题报告

Description
Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].


Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].


Challenge
O(log n) time.

思路
二分法。
先找到第一个target，再找到最后一个target。


Code

public class Solution {
    /** 
     *@param A : an integer sorted array
     *@param target :  an integer to be inserted
     *return : a list of length 2, [index1, index2]
     */
    public int[] searchRange(int[] A, int target) {
        // write your code here
        
        if (A == null || A.length == 0) {
            return new int[] {-1, -1};
        }
        
        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] >= target) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        
        int first = 0;
        if (A[start] == target) {
            first = start;
        }
        else if (A[end] == target){
            first = end;
        }
        else {
            return new int[] {-1 -1};
        }
        
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] <= target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        
        int last = first;
        if (A[end] == target) {
            last = end;
        }
        else if (A[start] == target) {
            last = start;
        }
        return new int[] {first, last};
    }
}