Missing digits from 0 to n-1

package main

/**
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

*/

// 直观思维
func missingNumber(nums []int) int {
	if nums[0] != 0 {
		return 0
	}
	for i := 1; i < len(nums); i++ {
		if nums[i]-nums[i-1] == 2 {
			return nums[i] - 1
		}
	}
	return len(nums)
}

// nums[i] == i 遍历思路
func missingNumber1(nums []int) int {
	for i := 0; i < len(nums); i++ {
		if nums[i] != i {
			return i
		}
	}
	return len(nums)
}

// 二分查找
func missingNumber2(nums []int) int {
	left, right := 0, len(nums)-1
	for left <= right {
		m := left + (right-left)/2
		if nums[m] == m { // 说明在右半部分
			left = m + 1
		} else {
			right = m - 1
		}
	}
	return left
}

package main

import (
	"testing"

	"github.com/bmizerany/assert"
)

func TestMissNumber(t *testing.T) {
	t.Run("首", func(t *testing.T) {
		m := missingNumber([]int{1, 2, 3})
		assert.Equal(t, 0, m, 0)
	})
	t.Run("尾", func(t *testing.T) {
		m := missingNumber([]int{0, 1, 2})
		assert.Equal(t, 3, m, 3)
	})
	t.Run("中间", func(t *testing.T) {
		m := missingNumber([]int{0, 1, 3})
		assert.Equal(t, 2, m, 2)
	})
}

func TestMissNumber1(t *testing.T) {
	t.Run("首", func(t *testing.T) {
		m := missingNumber1([]int{1, 2, 3})
		assert.Equal(t, 0, m, 0)
	})
	t.Run("尾", func(t *testing.T) {
		m := missingNumber1([]int{0, 1, 2})
		assert.Equal(t, 3, m, 3)
	})
	t.Run("中间", func(t *testing.T) {
		m := missingNumber1([]int{0, 1, 3})
		assert.Equal(t, 2, m, 2)
	})
}

func TestMissNumber2(t *testing.T) {
	t.Run("首", func(t *testing.T) {
		m := missingNumber2([]int{1, 2, 3})
		assert.Equal(t, 0, m, 0)
	})
	t.Run("尾", func(t *testing.T) {
		m := missingNumber2([]int{0, 1, 2})
		assert.Equal(t, 3, m, 3)
	})
	t.Run("中间", func(t *testing.T) {
		m := missingNumber2([]int{0, 1, 3})
		assert.Equal(t, 2, m, 2)
	})
}