题目描述
给定一个按照升序排列的整数数组 nums,和一个目标值 target。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]。
进阶:
你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗?
示例
示例 1:
输入:nums = [5,7,7,8,8,10], target = 8
输出:[3,4]
来源:力扣(LeetCode)
链接:leetcode-cn.com/problems/fi…
实现
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* searchRange(int* nums, int numsSize, int target, int* returnSize){
int *result = (int *)malloc(sizeof(int) * 2);
*returnSize = 2;
result[0] = -1;
result[1] = -1;
if (numsSize == 0) {
return result;
}
if (numsSize == 1) {
if (nums[0] == target) {
result[0] = 0;
result[1] = 0;
}
return result;
}
int l = 0;
int r = numsSize - 1;
int mid = 0;
while (l <= r) {
mid = (l + r) / 2;
if (nums[mid] > target) {
r = mid - 1;
} else if (nums[mid] < target) {
l = mid + 1;
} else {
int left = mid;
int right = mid;
while ((left - 1) >= l && nums[left-1] == target) {
left--;
}
while ((right + 1) <= r && nums[right+1] == target) {
right++;
}
result[0] = left;
result[1] = right;
*returnSize = 2;
return result;
}
}
return result;
}