剑指 offer 53 ii. 0～n 1中缺失的数字

剑指 Offer 53 - II. 0～n-1中缺失的数字

一个长度为n-1的递增排序数组中的所有数字都是唯一的，并且每个数字都在范围0～n-1之内。在范围0～n-1内的n个数字中有且只有一个数字不在该数组中，请找出这个数字。

示例 1:

输入: [0,1,3]
输出: 2

示例 2:

输入: [0,1,2,3,4,5,6,7,9]
输出: 8

限制：

• 1 <= 数组长度 <= 10000

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/que-shi-de-shu-zi-lcof 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution 1

二分查找

class Solution {
    public int missingNumber(int[] nums) {
        int index = binarySearchMissing(nums, 0, nums.length - 1);
        if (index == -1)
            return nums.length;
        return index;
    }

    private int binarySearchMissing(int[] nums, int start, int end) {
        int mid = 0;
        while (start <= end) {
            mid = (start + end) / 2;
            if (nums[mid] != mid && (mid == 0 || nums[mid - 1] == mid - 1))
                return mid;
            else if (nums[mid] == mid)
                start = mid + 1;
            else
                end = mid - 1;
        }
        return -1;
    }
}

Solution 2

cpp

#include <vector>
using std::vector;

class Solution
{
public:
    int missingNumber(vector<int> &nums)
    {
        int left = 0;
        int right = nums.size() - 1;
        int mid;
        while (left <= right)
        {
            mid = (left + right) / 2;
            if (nums[mid] == mid)
                left = mid + 1;
            else // if (nums[mid] == mid + 1)
            {
                if (mid == 0)
                    return 0;
                if (nums[mid - 1] == mid - 1)
                    return mid;
                right = mid - 1;
            }
        }
        return left;
    }
};