Search for a Range

来源：互联网发布：西门子200plc编程实例编辑：程序博客网时间：2024/06/06 02:14

原题：

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].

解题思路：

二分查找

代码1：

public int[] searchRange(int[] A, int target) {        int[] result = new int[2];        result[0] = -1;        result[1] = -1;        if(A==null || A.length<1) {        return result;        }        result[0] = findLeft(A, target);        result[1] = findRight(A, target);                return result;    }private int findLeft(int[] A, int target) {int result = 0;int left = 0;        int right = A.length-1;        int middle;        while(left <= right) {        if( A[left]==target&&(left==0||A[left-1]!=target) ) {        return left;        }        middle = left + (right-left)/2;        if(A[middle]==target) {        if( (middle==0||A[middle-1]!=target) ) {            return middle;            } else {            right = middle-1;            }        } else if(A[middle]<target) {        left = middle+1;        } else {        right = middle-1;        }                }return result;}private int findRight(int[] A, int target) {int result = 0;int left = 0;        int right = A.length-1;        int middle;        while(left <= right) {        if( A[right]==target&&(right==A.length-1 || A[right+1]!=target)) {        return result;        }        middle = left + (right-left)/2;        if(A[middle]==target) {        if( (middle==A.length-1||A[middle+1]!=target) ) {            return middle;            } else {            left = middle+1;            }        } else if(A[middle]<target) {        left = middle+1;        } else {        right = middle-1;        }                }return result;}

另一种写法：

public int[] searchRange(int[] A, int target) {        int[] result = new int[2];        result[0] = -1;        result[1] = -1;        if(A==null || A.length<1) {        return result;        }        int left = 0;        int right = A.length-1;        int middle;        while(left<=right) {        middle = left + (right-left)/2;        if(A[middle]<target) {        left = middle+1;        } else if(A[middle]>target) {        right = middle-1;        } else {        result[0] = lowerBound(A, left, middle-1, target);        result[1] = upperBound(A, middle+1, right, target);        return result;        }        }                return result;    }private int lowerBound(int[] A, int left, int right, int target) {while(left<=right) {int middle = left + (right-left)/2;if(A[middle]<target) {left = middle+1;} else {right = middle-1;}}return right+1;}private int upperBound(int[] A, int left, int right, int target) {while(left<=right) {int middle = left + (right-left)/2;if(A[middle]<=target) {left = middle+1;} else {right = middle-1;}}return left-1;}

0 0