Leetcode buckle classic problem -- 4. Find the median of two positively ordered arrays

4. Find the median of two positive arrays

Title Description ：

 1. Merge to find the median

Ideas ： The solution we thought of at the beginning of getting this problem must be violent solution , Directly combine the two arrays to find the median .

When the merged array is even , Then put the length/2 Elements and number length-1/2 Elements are averaged ; If it's an odd number , Then go directly to No length/2 Elements ;

The code is as follows ：

    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int[] temp = new int[nums1.length+nums2.length];
        int i = 0,j = 0,t = 0;
        while (i < nums1.length && j < nums2.length){
            if (nums1[i] < nums2[j]){
                temp[t] = nums1[i];
                i++;
                t++;
            }else {
                temp[t] = nums2[j];
                j++;
                t++;
            }
        }
        while ( i < nums1.length){
            temp[t] = nums1[i];
            i++;
            t++;
        }
        while (j< nums2.length){
            temp[t] = nums2[j];
            j++;
            t++;
        }
        double t1 = temp[temp.length/2];
        double t2 = temp[(temp.length-1)/2];
        if (temp.length % 2 == 0)
            return  (t1+t2) / 2.0;
        else 
            return t2;        
    }

Time complexity ： Traversed all arrays O(m+n)

Spatial complexity ： A new array is defined ,O(m+n)

2. Do not merge ( Traverse to the median )

Ideas ： The first method is to merge the arrays directly , Then find the median , In fact, when we think about it carefully, we know that we can not merge , Directly traverse to the median and directly return ; But now the problem is , When the number is odd, we can directly return to the median , But what about even numbers ？

If it's even , We need to record the number length/2 Elements and number length-1/2 Elements , We can introduce two variables ,left、right, After each traversal right Value is assigned to left(left = right), So when we traverse to the last number (length/2), You will find that just left The value of ,left= The first length-1/2 Elements ,right= The first length-1/2 Elements , So we can judge later ;

Let's look at the code ：

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length,n = nums2.length;
        int len = m + n;
        int left = 0,right = 0;
        int start1 = 0,start2 = 0;   // Start subscript 
        for (int i = 0; i <= len/2; i++) {
            left = right;
            if (start1 < m && (start2 >= n || nums1[start1] < nums2[start2]))
                right = nums1[start1++];
            else
                right = nums2[start2++];
        }
        if (len % 2 == 0)   // Combine to an even number 
            return (left+right)/2.0;
        else
            return right;
    }

Time complexity ： Ergodic len/2+1 Time , Still O(m+n)

Spatial complexity ：O(1)

summary ： Although it is positioned as a difficult problem , But the topic logic is not complicated , Just draw a picture and think about it