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LintCode Search For a Range (Binary Search)

时间:2016-05-03 15:57:22      阅读:170      评论:0      收藏:0      [点我收藏+]

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Binary Search模板: mid 和 target 指针比较,left/ right 和 target 比较。

 

循环终止条件: 最后剩两数比较(while(left + 1 < right))。

循环结束后根据要求检查最后两个数(left/ right 和 target 比较)。

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) {
        int[] array = new int[2];
        array[0] = -1;
        array[1] = -1;
        if(A == null || A.length == 0) return array;
        
        int left = 0; int right = A.length - 1;
        while(left + 1 < right){
            int mid = (left + right) / 2;
            if(A[mid] == target){
                right = mid;
            }
            else if(A[mid] < target){
                left = mid;
            }
            else if(A[mid] > target){
                right = mid;
            }
        }
        if(A[left] == target){
            array[0] = left;
        }
        else if(A[right] == target){
            array[0] = right;
        }
        else array[0] = -1;
        
        left = 0; right = A.length - 1;
        while(left + 1 < right){
            int mid = (left + right) / 2;
            if(A[mid] == target){
                left = mid;
            }
            else if(A[mid] < target){
                left = mid;
            }
            else if(A[mid] > target){
                right = mid;
            }
        }
        if(A[right] == target){
            array[1] = right;
        }
        else if(A[left] == target){
            array[1] = left;
        }
        else array[1] = -1;
        return array;
    }
}

 

LintCode Search For a Range (Binary Search)

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原文地址:http://www.cnblogs.com/LittleAlex/p/5455126.html

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