[leetcode] 4. 寻找两个有序数组的中位数

官方题解:https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xun-zhao-liang-ge-you-xu-shu-zu-de-zhong-wei-shu-b/

class Solution {
public:
    double findMedianSortedArrays(vector<int>& A, vector<int>& B) {
        int m = A.size();
        int n = B.size();
        if(m > n)
        {
            vector<int> temp = A;
            A = B;
            B = temp;
            m = A.size();
            n = B.size();
        }
        int iMin = 0, iMax = m, halfLen = (n + m + 1)/2;
        while(iMin <= iMax)
        {
            int i = (iMin+iMax)/2;  //0 <= i <= m;
            int j = halfLen - i;
            if(i < iMax && B[j-1] > A[i])
            {//i < m ==> j > 0
            //并且m < n ==> j <= n  
               iMin = i + 1;        //too small
            }
            else if(i > iMin && A[i-1] > B[j])
            {//i > 0 ==> j < n
            //并且m < n ==> j >= 0 
                iMax = i - 1;       //too big
            }
            else
            {
                double maxLeft = 0;
                if(i == 0){ maxLeft = B[j-1]; }
                else if(j == 0) { maxLeft = A[i-1]; }
                else {  maxLeft = 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 = min(A[i], B[j]); }

                return (maxLeft + minRight)/2.0;
            }
        }
        return 0.0;
    }
};

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m <= n的作用：

            if(i < iMax && B[j-1] > A[i])
            {//i < m ==> j > 0 
            //并且m < n ==> j <= n  
               iMin = i + 1;        //too small
            }

由i < m ==> j > 0，但是，这个时候j会大于n吗？
不会，因为i是大于等于0小于等于m的，由m <= n（一开始保证了) 可以推出j < = n,因为i越小,j就越大,i == 0时,j最大，j = (m + n + 1) /2 - 0 <= (2n + 1)/2 <= n(int运算)

			else if(i > iMin && A[i-1] > B[j])
            {//i > 0 ==> j < n
            //并且m < n ==> j >= 0 
                iMax = i - 1;       //too big
            }

这个也是同理,由i > 0 ==> j < n,由m < n ==> j >= 0，因为i越大,j就越小,i == m时,j最大，j = (m + n + 1) /2 - m >= (n - m + 1)/2 >= 0(int运算,m <= n)