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34. Find First and Last Position of Element in Sorted Array
Given an array of integers nums sorted in ascending order, 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].
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]
思路分析
可以直接使用二分查找，搜索两次。
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public static int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return new int[]{-1, -1};
}
// 搜索左边界
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] < target) {
left = mid + 1;
} else if (target < nums[mid]) {
right = mid - 1;
} else if (nums[mid] == target) {
right = mid - 1; // 注意这里的处理
}
}
if (left == nums.length) {
return new int[]{-1, -1};
}
int i1 = -1;
if (nums[left] == target) {
i1 = left;
} else {
return new int[]{-1, -1};
}
// 搜索右边界
left = 0;
right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] < target) {
left = mid + 1;
} else if (target < nums[mid]) {
right = mid - 1;
} else if (nums[mid] == target) {
left = mid + 1; // 注意这里的处理
}
}
if (left - 1 < 0) {
return new int[]{-1, -1};
}
int i2 = -1;
if (nums[left - 1] == target) {
i2 = left - 1;
} else {
return new int[]{-1, -1};
}
return new int[]{i1, i2};
}