//Given an array of integers nums sorted in non-decreasing order, find the start
//ing and ending position of a given target value.
//
// If target is not found in the array, return [-1, -1].
//
// You must write an algorithm with O(log n) runtime complexity.
//
//
// Example 1:
// Input: nums = [5,7,7,8,8,10], target = 8
//Output: [3,4]
// Example 2:
// Input: nums = [5,7,7,8,8,10], target = 6
//Output: [-1,-1]
// Example 3:
// Input: nums = [], target = 0
//Output: [-1,-1]
//
//
// Constraints:
//
//
// 0 <= nums.length <= 105
// -109 <= nums[i] <= 109
// nums is a non-decreasing array.
// -109 <= target <= 109
//
// Related Topics Array Binary Search
// 👍 11383 👎 306
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = new int[2];
result[0] = searchRangeFunc(nums, target, true);
result[1] = searchRangeFunc(nums, target, false);
return result;
}
private int searchRangeFunc(int[] nums, int target, boolean left) {
int from = 0, end = nums.length - 1;
while (from <= end) {
if (from == end) {
if (nums[from] == target) {
return from;
} else {
return -1;
}
}
if (from + 1 == end) {
if (nums[from] == target && nums[end] == target) {
if (left) {
return from;
} else {
return end;
}
}
if (nums[from] == target) {
return from;
} else if (nums[end] == target) {
return end;
} else {
return -1;
}
}
int mid = from + (end - from) / 2;
if (nums[mid] < target) {
from = mid;
} else if (nums[mid] > target) {
end = mid;
} else {
if (left) {
end = mid;
} else {
from = mid;
}
}
}
return -1;
}
}
//leetcode submit region end(Prohibit modification and deletion)