Topic: 34. Find the first and last position of an element in a sorted array
Given an integer array nums arranged in ascending order, and a target value target. Find the start and end positions of the given target value in the array.
If the target value target does not exist in the array, return [-1, -1].
answer:
Method 1: The intuitive idea is definitely to traverse from the front to the back. Two variables are used to record the first and last subscripts encountered, but the time complexity of this method is O(n)O(n), and the condition of the ascending order of the array is not used.
int* searchRange(int* nums, int numsSize, int target, int* returnSize)
{
// 开辟两个int大小的空间放索引
int* res = (int*)malloc(sizeof(int) * 2);
// 判断是否找到target
int flag = 0;
int i = 0;
while (i < numsSize)
{
// 找到target，flag置1
if (nums[i] == target)
{
flag = 1;
break;
}
i++;
}
// 没找到target，返回[-1,-1]
if (flag == 0)
{
res[0] = -1;
res[1] = -1;
*returnSize=2;
return res;
}
// 程序走到这里，一定存在索引
// 先记录第一个索引位置
res[0] = i;
// 寻找结束位置
while (i < numsSize && nums[i] == target)
{
i++;
}
// 再记录结束索引位置
res[1] = i - 1;
*returnSize=2;
return res;
}
Method 2: Since the array has been sorted, the entire array is monotonically increasing. We can use the dichotomy to speed up the search process.
In fact, what we are looking for is "the first position equal to the target" and "the first position greater than the target minus one" in the array.
// 找出给定目标值在数组中的开始位置
int findFitstPosition(int* nums, int numsSize, int target)
{
int size = numsSize;
int left = 0;
int right = size - 1;
while (left < right)
{
// 取下界
int mid = left + (right - left) / 2;
if (nums[mid] < target)
left = mid + 1;
else if (nums[mid] == target)
// 即使nums[mid] == target，也要right = mid
// 使得下标向target的第一个位置靠拢
right = mid;
else
// nums[mid] > target
right = mid - 1;
}
// left == right
if (nums[left] != target)
return -1;
return left;
}
// 找出给定目标值在数组中的结束位置
int findLastPosition(int* nums, int numsSize, int target)
{
int size = numsSize;
int left = 0;
int right = size - 1;
while (left < right)
{
// 取上界
int mid = left + (right - left + 1) / 2;
if (nums[mid] > target)
right = mid - 1;
else if (nums[mid] == target)
// 使得下标向target的最后一个位置靠拢
left = mid;
else
// nums[mid] < target
left = mid + 1;
}
// left == right
if (nums[left] != target)
return -1;
return left;
}
// 在排序数组中查找元素的第一个和最后一个位置
int* searchRange(int* nums, int numsSize, int target, int* returnSize)
{
int* ans=(int*)malloc(sizeof(int)*2);
int size = numsSize;
// 整数数组为空，直接返回[-1,-1]
if (size == 0)
{
ans[0]=-1;
ans[1]=-1;
*returnSize=2;
return ans;
}
// 没有找到target，，直接返回[-1,-1]
int fitstPosition = findFitstPosition(nums, numsSize, target);
if (fitstPosition == -1)
{
ans[0]=-1;
ans[1]=-1;
*returnSize=2;
return ans;
}
int lastPosition = findLastPosition(nums, numsSize, target);
ans[0]=fitstPosition;
ans[1]=lastPosition;
*returnSize=2;
return ans;
}
// 找出给定目标值在数组中的开始位置
int findFitstPosition(int* nums, int numsSize, int target)
{
if(numsSize==0)
return -1;
int size = numsSize;
int left = 0;
int right = size - 1;
while (left < right)
{
// 取下界
int mid = left + (right - left) / 2;
if (nums[mid] < target)
left = mid + 1;
else if (nums[mid] == target)
// 即使nums[mid] == target，也要right = mid
// 使得下标向target的第一个位置靠拢
right = mid;
else
// nums[mid] > target
right = mid - 1;
}
// left == right
if (nums[left] != target)
return -1;
return left;
}
// 找出给定目标值在数组中的结束位置
int findLastPosition(int* nums, int numsSize, int target)
{
if(numsSize==0)
return -1;
int size = numsSize;
int left = 0;
int right = size - 1;
while (left < right)
{
// 取上界
int mid = left + (right - left + 1) / 2;
if (nums[mid] > target)
right = mid - 1;
else if (nums[mid] == target)
// 使得下标向target的最后一个位置靠拢
left = mid;
else
// nums[mid] < target
left = mid + 1;
}
// left == right
if (nums[left] != target)
return -1;
return left;
}
// 在排序数组中查找元素的第一个和最后一个位置
int* searchRange(int* nums, int numsSize, int target, int* returnSize)
{
int* ans=(int*)malloc(sizeof(int)*2);
int size = numsSize;
int fitstPosition = findFitstPosition(nums, numsSize, target);
// 整数数组为空 或者 没有找到target
// 直接返回[-1,-1]
// 两种情况放在一起的话，前面的两个函数要先排除size=0的情况
// 否则 right = size-1 < 0
// 代码复用，更加简洁
if (size == 0 || fitstPosition == -1)
{
ans[0]=-1;
ans[1]=-1;
*returnSize=2;
return ans;
}
int lastPosition = findLastPosition(nums, numsSize, target);
ans[0]=fitstPosition;
ans[1]=lastPosition;
*returnSize=2;
return ans;
}