分治法求兩個有序數組的中位數(LeedCode4)

分治法求兩個有序數組的中位數

算法步驟(基本原理是獲取第k小的數)
   	先取兩個中間索引x_mid,y_mid;
    下面來比較
    如果x[x_mid]比較小,那麼就看x_mid最大是第幾小(記作m)
    ①m<k;
    則把x_mid及以前的全部刪除,同時變換k的值
    同時如果也要把y_mid及右邊的儘可能的刪除
    ②如果m==k
    則把x_mid以前的全部刪除並且保留x_mid
    同時如果也要把y_mid及右邊的儘可能的刪除
    同理如果y大也一樣
    如此遞歸,最後到遞歸出口爲k==1或者有數組爲空了

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if(nums2.length==0){
            if(nums1.length%2==0){
                return (double) (nums1[(nums1.length-1)/2]+nums1[(nums1.length-1)/2+1])/2.0;
            }else
                return (double)nums1[(nums1.length-1)/2];
        }else if(nums1.length==0){
            if(nums2.length%2==0){
                return (double) (nums2[(nums2.length-1)/2]+nums2[(nums2.length-1)/2+1])/2.0;
            }else
                return (double)nums2[(nums2.length-1)/2];
        }
        
        int i;
        if((nums1.length+nums2.length)%2==0){
            i=(nums1.length+nums2.length)/2;
            return (get_Middle_num(nums1,0,nums1.length-1,nums2,0,nums2.length-1,i)     
+get_Middle_num(nums1,0,nums1.length-1,nums2,0,nums2.length-1,i+1))/2.0;
        }
        else{
            i=(nums1.length+nums2.length)/2+1;
            return (double)get_Middle_num(nums1,0,nums1.length-1,nums2,0,nums2.length-1,i);
        }
    }
    int get_Middle_num(int x[],int left_x,int right_x,int y[],int left_y,int right_y,int k){*/
        if(left_x>right_x){
            return y[left_y+k-1];
        }else if(left_y>right_y){
            return x[left_x+k-1];
        }else if(k==1){
                return x[left_x]>y[left_y]?y[left_y]:x[left_x];
        }else{
            int x_mid = (left_x+right_x)/2;
            int y_mid = (left_y+right_y)/2;
            int num = (x_mid-left_x)+(y_mid-left_y);
            //注意num+1代表含義:x_mid可以到達的最大第幾小
            // num+2代表含義:y_mid可以到達的最小第幾小
            if(x[x_mid]<=y[y_mid]){
                //接下來我們開始鎖定x左邊界和y的右邊界
                if(num+2==k){
                    right_y=y_mid;
                }else if(num+2>k)
                    right_y=y_mid-1;

                if(num+1<k){
                    k-=x_mid+1-left_x;
                    left_x=x_mid+1;
                }
                else if(num+1==k){
                    k-=x_mid-left_x;
                    left_x=x_mid;
                }
               return get_Middle_num(x,left_x, right_x,y,left_y,right_y,k);
            }else{
                 //接下來我們開始鎖定x左邊界和y的右邊界
                if(num+2==k){
                    right_x=x_mid;
                }else if(num+2>k)
                    right_x=x_mid-1;

                if(num+1<k){
                    k-=y_mid+1-left_y;
                    left_y=y_mid+1;
                }
                else if(num+1==k){
                    k-=y_mid-left_y;
                    left_y=y_mid;
                }
               return get_Middle_num(x,left_x, right_x,y,left_y,right_y,k);
            }
        }
    }
}

算法複雜度分析:
T(n)=T(n/4)+O(1)

O(n）=logn

最後附上測試用例來供讀者使用

    @Test
    public  void Test(){
        //[76,89,104,30823,31070,31259,31324,31714,31971,320331]
        //[122,2919,2941,17754,17781,17830,17874,18019,18023,18130]
        //[2634,2764,2801,30823,31070,31259,32319,32350,32408,32475,32681,32701,32764]
        //[122,32605,32641,32716,32757]
        int x[]={2634,2764,2801,30823,31070,31259,32319,32350,32408,32475,32681,32701,32764};
        int y[]={122,32605,32641,32716,32757};
        System.out.println(get_Middle_num(x,0,x.length-1,y,0,y.length-1,1));
        System.out.println((x.length+y.length));
    }