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26章ソートされた配列から重複を削除する

容易

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Python Solution Using an Iterative approach

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:

        for i in range(len(nums)):
            if nums[i] >= target:
                return i
                break
            elif nums[i] > target and target < nums[i + 1]:
                return i+1
                break
            elif target > nums[-1]:
                return len(nums)
            elif target < nums[0]:
                return 0
                break

        return i

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        n = len(nums)

        start = 0
        end = n - 1

        while start <= end:
            mid = (start + end) // 2

            if nums[mid] == target:
                return mid
            elif nums[mid] < target:
                start = mid + 1
            else:
                end = mid - 1

        return end + 1