解題思路： 雙指針/二分查找法（官網(wǎng)）
如果數(shù)組中不存在目標(biāo)值 target，返回?[-1, -1]蝗羊。
進(jìn)階：
你可以設(shè)計(jì)并實(shí)現(xiàn)時(shí)間復(fù)雜度為?O(log n)?的算法解決此問(wèn)題嗎固棚？
示例 1：
輸入：nums = [5,7,7,8,8,10], target = 8
輸出：[3,4]
示例?2：
輸入：nums = [5,7,7,8,8,10], target = 6
輸出：[-1,-1]
示例 3：
輸入：nums = [], target = 0
輸出：[-1,-1]
提示：
0 <= nums.length <= 105
-109?<= nums[i]?<= 109
nums?是一個(gè)非遞減數(shù)組
-109?<= target?<= 109
來(lái)源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
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代碼實(shí)現(xiàn)：
class?Solution?{
????public?int[]?searchRange(int[]?nums,?int?target)?{
????????if?(nums?==?null?||?nums.length?==?0)?{
????????????return?new?int[]{-1,?-1};
????????}
????????int?len?=?nums.length;
????????int?head?=?0;?int?tail?=?len?-?1;
????????boolean?foundHead?=?false;
????????boolean?foundTail?=?false;
????????while?(head?<=?tail)?{
????????????if?(!foundHead)?{
????????????????if?(nums[head]?==?target)?{
????????????????????foundHead?=?true;
????????????????}?else?{
????????????????????head?++;
????????????????}
????????????}
????????????if?(!foundTail)?{
????????????????if?(nums[tail]?==?target)?{
????????????????????foundTail?=?true;
????????????????}?else?{
????????????????????tail?--;
????????????????}
????????????}
????????????if?(foundHead?&&?foundTail)?{
????????????????break;
????????????}
????????}
????????if?(foundHead?&&?foundTail)?{
????????????return?new?int[]{head,?tail};
????????}?else?{
????????????return?new?int[]{-1,?-1};
????????}
????}
}
