leetcode4

给定两个大小分别为 m 和 n 的正序（从小到大）数组 num1 和 num2。请你找出并返回这两个正序数组的 中位数
当数组为偶数个时，中位数为两位数字的平均值

        func findMedianSortedArrays(_ num1: [Int], _ num2: [Int]) -> Double{
            guard num1.count > 0 || num2.count > 0 else {
                return -1.0
            }
            var middle = 0.0  //中位数, 当数组为偶数个时，中位数为两位数字的平均值
            if num1.count == 0 {  //num1为空数组
                let mid = num2.count / 2
                if num2.count%2 == 0 {  //偶数个，取数组最后两个的平均值
                    middle = Double(num2[mid] + num2[mid-1])/2.0
                }else{
                    middle = Double(num2[mid])
                }
                return middle
            }else if num2.count == 0 {  //num2为空数组
                let mid = num1.count / 2
                if num1.count%2 == 0 {  //偶数个，取数组最后两个的平均值
                    middle = Double(num1[mid] + num1[mid-1])/2.0
                }else{
                    middle = Double(num1[mid])
                }
                return middle
            }
            
            var merg = [Int]()
            var left=0, right=0
            while left < num1.count, right < num2.count {
                if merg.count > (num2.count+num1.count)/2 {  //因为是取中位数，因此舍弃后续需排序的元素，直接结束
                    break
                }
                if num1[left] < num2[right] { //num1的第一个元素小于num2的第一个元素
                    merg.append(num1[left])
                    left += 1
                }else{
                    merg.append(num2[right])
                    right += 1
                }
            }
            let miss = (num2.count+num1.count)/2+1 - merg.count   //数组达到标准还差多少个
            if miss > 0 {//两个数组长度偏差过大，新数组还不超总长度的一半
                if num1.count > num2.count {  //num1数组更长
                    let has = merg.count-num2.count  //num1数组中已经加入到merg数组中元素的个数
                    merg += num1[has..<(has+miss)]
                }else if num2.count == num1.count { //两个数组一样长，此时可能在num1取值，也可能在num2取值. 总的原则是在大数的数组中取值
                    if num1.last! > num2.last! {
                        let has = merg.count-num2.count  //num1数组中已经加入到merg数组中元素的个数
                        merg += num1[has..<(has+miss)]
                    }else{
                        let has = merg.count-num1.count  //num1数组中已经加入到merg数组中元素的个数
                        merg += num2[has..<(has+miss)]
                    }
                }
                else{
                    let has = merg.count-num1.count  //num1数组中已经加入到merg数组中元素的个数
                    merg += num2[has..<(has+miss)]
                }
            }
            if (num2.count+num1.count)%2 == 0 {  //偶数个，取数组最后两个的平均值
                middle = Double(merg[merg.count-1] + merg[merg.count-2])/2.0
            }else{
                middle = Double(merg.last!)
            }
            
            return middle
        }
        print(findMedianSortedArrays([100001], [100000]))