这题如果分别写两个二分查找，还比较简单；如果想写同一个函数做两次二分，就有多一些边界情况要考虑：
class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length == 0) return new int[]{-1, -1};
int first = findFirstGreaterOrEqualsTo(nums, target, 0);
if (first == nums.length || nums[first] != target) return new int[]{-1, -1};
int last = findFirstGreaterOrEqualsTo(nums, target + 1, first + 1) - 1;
return new int[]{first, last};
}
int findFirstGreaterOrEqualsTo(int[] nums, int target, int start) {
int l = start, r = nums.length;
// System.out.println("find first...");
while (l < r) {
// System.out.println("l: " + l + " r: " + r);
int mid = l + (r - l) / 2, midVal = nums[mid];
if (midVal >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}
