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Search for a Range

描述

Given a sorted array of integers, 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].

For example, Given [5, 7, 7, 8, 8, 10] and target value 8, return [3, 4].

分析

已经排好了序,用二分查找。

重新实现 lower_bound 和 upper_bound

// Search for a Range
// 重新实现 lower_bound 和 upper_bound
// 时间复杂度O(logn),空间复杂度O(1)
public class Solution {
public int[] searchRange(int[] nums, int target) {
int lower = lower_bound(nums, 0, nums.length, target);
int upper = upper_bound(nums, 0, nums.length, target);

if (lower == nums.length || nums[lower] != target)
return new int[]{-1, -1};
else
return new int[]{lower, upper-1};
}

int lower_bound (int[] A, int first, int last, int target) {
while (first != last) {
int mid = first + (last - first) / 2;
if (target > A[mid]) first = ++mid;
else last = mid;
}

return first;
}

int upper_bound (int[] A, int first, int last, int target) {
while (first != last) {
int mid = first + (last - first) / 2;
if (target >= A[mid]) first = ++mid; // 与 lower_bound 仅此不同
else last = mid;
}

return first;
}
}

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