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]
提示:
0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums
是一个非递减数组-109 <= target <= 109
代码
核心逻辑:针对有重复目标元素的数组,采用二分查找的变体。
- 找左边界:不断推动
right
向左移动 - 找右边界:不断推动
left
向右移动
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int left = findLeftPosition(nums, target);
int right = findRightPosition(nums, target);
return {
left, right};
}
int findLeftPosition(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int result = -1;
while (left <= right) {
int middle = (left + right) / 2;
if (nums[middle] >= target) {
right = middle - 1;
if (nums[middle] == target) {
result = middle;
}
}
else {
left = middle + 1;
}
}
return result;
}
int findRightPosition(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int result = -1;
while (left <= right) {
int middle = (left + right) / 2;
if (nums[middle] <= target) {
left = middle + 1;
if (nums[middle] == target) {
result = middle;
}
}
else {
right = middle - 1;
}
}
return result;
}
};