题目
给你一个按照非递减顺序排列的整数数组 nums,和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。 如果数组中不存在目标值 target,返回 [-1, -1]。 你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。
示例1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]
示例 3:
输入: nums = [], target = 0
输出: [-1,-1]
解题思路
寻找 target 在数组里的左右边界,一共有三种情况:
- target在数组范围的右边或者左边,此时应该返回 [-1, -1]
- target在数组范围中,且数组中不存在 target,如示例2,此时应该返回 [-1, -1]
- target在数组范围中,且数组中存在 target,如示例1,此时应该返回正确的边界值
寻找左右边界
先分别去获取target的左右边界,采用二分法,左闭右闭 寻找右边界 如果要寻找右边界,那就是尽可能的去寻找最右边的target
- 当 nums[mid] > target 时,right = mid - 1;
- 当 nums[mid] < target 时,left = mid + 1,rightBorder = left;
- 当 nums[mid] = target 时,因为需要尽可能的去寻找最右边的target,所以在这种情况下,还需要继续寻找,left = mid + 1,rightBorder = left。
综上,代码实现如下:
public static int getRightBorder(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int rightBorder = -2;
while (left <= right) {
int mid = left + ((right - left) / 2);
if (nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
rightBorder = left;
}
}
return rightBorder;
}
寻找左边界 和寻找右边界类似,寻找左边界,就是尽可能的去寻找最左边的target
- 当 nums[mid] < target 时,left = mid + 1;
- 当 nums[mid] > target 时,right = mid - 1,leftBorder = right;
- 当 nums[mid] = target 时,因为需要尽可能的去寻找最左边的target,所以在这种情况下,还需要继续寻找,right = mid - 1,leftBorder = right;
综上,代码实现如下:
public static int getLeftBorder(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int leftBorder = -2;
while (left <= right) {
int mid = left + ((right - left) / 2);
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
leftBorder = right;
}
}
return leftBorder;
}
处理三种情况 代码实现如下:
public static int[] searchRange(int[] nums, int target) {
int leftBorder = getLeftBorder(nums, target);
int rightBorder = getRightBorder(nums, target);
// 情况1
if (leftBorder == -2 || rightBorder == -2) {
return new int[]{-1, -1};
}
// 情况三
if (rightBorder - leftBorder > 1) {
return new int[]{leftBorder + 1, rightBorder - 1};
}
// 情况二
return new int[]{-1, -1};
}
代码实现
public class Question34 {
public static int[] searchRange(int[] nums, int target) {
int leftBorder = getLeftBorder(nums, target);
int rightBorder = getRightBorder(nums, target);
if (leftBorder == -2 || rightBorder == -2) {
return new int[]{-1, -1};
}
if (rightBorder - leftBorder > 1) {
return new int[]{leftBorder + 1, rightBorder - 1};
}
return new int[]{-1, -1};
}
public static int getRightBorder(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int rightBorder = -2;
while (left <= right) {
int mid = left + ((right - left) / 2);
if (nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
rightBorder = left;
}
}
return rightBorder;
}
public static int getLeftBorder(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int leftBorder = -2;
while (left <= right) {
int mid = left + ((right - left) / 2);
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
leftBorder = right;
}
}
return leftBorder;
}
public static String integerArrayToString(int[] nums, int length) {
if (length == 0) {
return "[]";
}
String result = "";
for (int index = 0; index < length; index++) {
int number = nums[index];
result += Integer.toString(number) + ", ";
}
return "[" + result.substring(0, result.length() - 2) + "]";
}
public static String integerArrayToString(int[] nums) {
return integerArrayToString(nums, nums.length);
}
public static void main(String[] args) {
int[] array = new int[]{-1, 0, 3, 3, 5, 9, 12};
System.out.println(integerArrayToString(searchRange(array, 3)));
}
}