个人记录-LeetCode 34. Search for a Range

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问题：
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].

代码示例：

1、这个题目没什么难度，从复杂度来看，二分法即可。

public class Solution {    public int[] searchRange(int[] nums, int target) {        if (nums == null || nums.length < 1) {            return new int[] {-1, -1};        }        if (nums.length < 2) {            if (nums[0] == target) {                return new int[] {0, 0};            } else {                return new int[] {-1, -1};            }        }        Set<Integer> result = new TreeSet<>();        searchResult(nums, 0, nums.length-1, target, result);        if (result.isEmpty()) {            return new int[] {-1, -1};        } else {            Object[] temp = result.toArray();            return new int[] {(int)temp[0], (int)temp[temp.length-1]};        }    }    private void searchResult(int[] nums, int begin, int end, int target, Set<Integer> result) {        if (begin > end) {            return;        }        int middle = (begin + end) / 2;        if (nums[middle] == target) {            result.add(middle);        }        if (nums[middle] >= target) {            if (nums[begin] <= target) {                searchResult(nums, begin, middle - 1, target, result);            }        }        if (nums[middle] <= target) {            if (nums[end] >= target) {                searchResult(nums, middle + 1, end, target, result);            }        }    }}

2、二分法反复递归，比较耗时，改进的方法：

public class Solution {    public int[] searchRange(int[] nums, int target) {        if (nums == null || nums.length < 1) {            return new int[] {-1, -1};        }        if (nums.length < 2) {            if (nums[0] == target) {                return new int[] {0, 0};            } else {                return new int[] {-1, -1};            }        }        int low = 0;        int high = nums.length - 1;        boolean flag = false;        while (low <= high) {            int middle = (low + high) / 2;            if (nums[middle] == target) {                flag = true;                low = middle;                while (low > 0 && nums[low-1] == target) {                    --low;                }                high = middle;                while (high < nums.length-1 && nums[high+1] == target) {                    ++high;                }                break;            } else if (nums[middle] > target){                high = middle - 1;            } else {                low = middle + 1;            }        }        if (flag) {            return new int[] {low, high};        } else {            return new int[] {-1, -1};        }    }}

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