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Search for a Range

时间:2016-08-11 00:49:59      阅读:164      评论:0      收藏:0      [点我收藏+]

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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].

 

 分析:

这道题只需要分别对起点和终点用二分查找就可以啦~

 

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) {
        if (A.length == 0) {
            return new int[]{-1, -1};
        }
        int[] result = new int[2];
        int start, end;
        
        //search the first element
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                end = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if(A[start] == target) {
            result[0] = start;
        } else if (A[end] == target){ 
            result[0] = end;
        } else {
            result[0] = result[1] = -1;
            return result;
        }
            
        //search the second element
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                start = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if(A[end] == target) {
            result[1] = end;
        } else if (A[start] == target){ 
            result[1] = start;
        } else {
            result[0] = result[1] = -1;
            return result;
        }
        return result;
    }
}

 

Search for a Range

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原文地址:http://www.cnblogs.com/iwangzheng/p/5759333.html

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