记录python2和3：list indices must be integers or slices, not float（查找两个有序数组的中位数遇到的问题）

起源是做查找两个有序数组的中位数用了python和java.
如果用下面的python代码就会出现list indices must be integers or slices, not float

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        m, n = len(nums1), len(nums2)
        if m>n:
            nums1, nums2, m, n = nums2, nums1, n, m
        if n==0:
            raise ValueError
        
        imin, imax, halflen = 0, m, (m+n+1)/2
        while imin<=imax:
            i = (imin + imax)/2
            j = halflen - i
            if i < m and nums2[j-1] > nums1[i]: imin = i+1
            elif i>0 and nums1[i-1]>nums2[j]: imax = i-1
            else:
                if i==0: maxleft = nums2[j-1]
                elif j==0: maxleft = nums1[i-1]
                else: maxleft = max(nums1[i-1], nums2[j-1])
                if (m+n)%2==1: return maxleft
                
                if i==m: minright = nums2[j]
                elif j==n: minright = nums1[i]
                else: minright = min(nums1[i], nums2[j])
                    
                return (maxleft+minright)/2.0
        
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原因是python2和3的不同，在python3中，使用“/”进行运算会得到一个浮点数，那么在用这个作为索引的时候，会出现比如nums1[0.5]这样的情况，显然是不能查找到的。所以在python3中，这里需要用“//”，得到的才是int。正确的python代码如下：

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        m, n = len(nums1), len(nums2)
        if m>n:
            nums1, nums2, m, n = nums2, nums1, n, m
        if n==0:
            raise ValueError
        
        imin, imax, halflen = 0, m, (m+n+1)//2
        while imin<=imax:
            i = (imin + imax)//2
            j = halflen - i
            if i < m and nums2[j-1] > nums1[i]: imin = i+1
            elif i>0 and nums1[i-1]>nums2[j]: imax = i-1
            else:
                if i==0: maxleft = nums2[j-1]
                elif j==0: maxleft = nums1[i-1]
                else: maxleft = max(nums1[i-1], nums2[j-1])
                if (m+n)%2==1: return maxleft
                
                if i==m: minright = nums2[j]
                elif j==n: minright = nums1[i]
                else: minright = min(nums1[i], nums2[j])
                    
                return (maxleft+minright)/2.0       
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java代码如下：

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m>n){
            int[] temp = A; A = B; B = temp;
            int tmp = m;  m = n; n = tmp;
        }
        
        int imin = 0, imax=m, halflen = (m+n+1)/2;
        while(imin<=imax){
            int i = (imin+imax)/2;
            int j = halflen-i;
            if(i<imax && B[j-1]>A[i]){
                imin = i+1;
            }
            else if(i>imin && A[i-1]>B[j]){
                imax = i-1;
            }
            else{
                int maxleft = 0;
                if(i==0) {maxleft = B[j-1];}
                else if(j==0) {maxleft=A[i-1];}
                else{ maxleft = Math.max(A[i-1],B[j-1]); }
                if( (m+n) % 2 == 1 ) {return maxleft;}
                
                int minright = 0;
                if (i == m) { minright = B[j]; }
                else if (j == n) { minright = A[i]; }
                else { minright = Math.min(B[j], A[i]); }

                return (maxleft + minright) / 2.0;
            }
        }
        return 0.0;
    }
}
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