4. 寻找两个正序数组的中位数

leetcode链接：
https://leetcode.cn/problems/median-of-two-sorted-arrays/description/

方案一

暴力求解

int[] flag = new int[nums1.length + nums2.length];

for (int i = 0; i < nums1.length; i++) {
	flag[i] = nums1[i];
}

for (int i = 0; i < nums2.length; i++) {
	flag[nums1.length + i] = nums2[i];
}

Arrays.sort(flag);

if ((flag.length % 2) == 1) {
	return flag[flag.length / 2];
} else {
	return (flag[flag.length / 2] + flag[(flag.length / 2) -1]) / 2.0;
}

执行结果：

• 2094/2094 cases passed (3 ms)
• Your runtime beats 21.6 % of java submissions
• Your memory usage beats 66.55 % of java submissions (42.2 MB)

几乎最优方案

https://leetcode.cn/problems/median-of-two-sorted-arrays/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-2/

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) { 
            return findMedianSortedArrays(B,A); // 保证 m <= n
        }
        int iMin = 0, iMax = m;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = (m + n + 1) / 2 - i;
            if (j != 0 && i != m && B[j-1] > A[i]){ // i 需要增大
                iMin = i + 1; 
            }
            else if (i != 0 && j != n && A[i-1] > B[j]) { // i 需要减小
                iMax = i - 1; 
            }
            else { // 达到要求，并且将边界条件列出来单独考虑
                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; } // 奇数的话不需要考虑右半部分

                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; //如果是偶数的话返回结果
            }
        }
        return 0.0;
    }
}

作者：windliang
链接：https://leetcode.cn/problems/median-of-two-sorted-arrays/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-2/
来源：力扣（LeetCode）
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