LeetCode 34 Search For A Range 二叉查找相关（二）

来源：互联网发布：大数据主要应用领域编辑：程序博客网时间：2024/06/07 07:05

题目：

https://leetcode.com/problems/search-for-a-range/

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].

最直接的想法就是先二叉查找到target的一个index，然后再往左往右分别查找边界，这样实际上有三遍二叉查找，代码如下：

public class Solution {    public int[] searchRange(int[] A, int target) {        int[] result = {-1, -1};        if (A == null || A.length == 0) {            return result;        }        int l = 0;        int r = A.length - 1;        boolean hasTarget = false;        int mid = (l + r) / 2;        while (l <= r) {            mid = (l + r) / 2;            if (A[mid] == target) {                hasTarget = true;                break;            } else if (A[mid] < target) {                l = mid + 1;            } else {                r = mid -1;            }        }        if (!hasTarget) {            return result;        }        // at this point, A[mid] = target, then need to find the range        // 1. find left range        l = 0;        r = mid;        result[1] = mid;        while (l <= r) {            mid = (l + r) / 2;            if (A[mid] == target) {                r = mid - 1;            } else {                // must be less than target                l = mid + 1;            }        }        result[0] = l;        // 2. find right range        l = result[1];        r = A.length - 1;        while (l <= r) {            mid = (l + r) / 2;            if (A[mid] == target) {                l = mid + 1;            } else {                // must be great than target                r = mid - 1;            }        }        result[1] = r;                return result;    }}

提交之后发现运行时间排在JAVA类的靠后，有没有优化的空间呢？

发现其实并不需要先找到一个Index之后再找左右的边界，可以一遍二叉查找左边界，再来一次查找右边界，代码如下：

public class Solution {    public int[] searchRange(int[] A, int target) {        int[] result = {-1, -1};        if (A == null || A.length == 0) {            return result;        }        int l = 0;        int r = A.length - 1;        // 1. find left range        while (l <= r) {            int mid = (l + r) / 2;            if (A[mid] >= target) {                r = mid - 1;            } else {                l = mid + 1;            }        }        if (l >= A.length || A[l] != target) {            // not found            return result;        }        result[0] = l;        // 2. find right range        r = A.length - 1;        while (l <= r) {            int mid = (l + r) / 2;            if (A[mid] == target) {                l = mid + 1;            } else {                // must be great than target                r = mid - 1;            }        }        result[1] = r;                return result;    }}

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