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  • 给定两个大小为 m 和 n 的有序数组 nums1 和 nums2。 请你找出这两个有序数组的中位数,并且要求算法的时间复杂度为 O(log(m + n))。 你可以假设 nums1 和 nums2 不会同时为空。

    class Solution {
        public double findMedianSortedArrays(int[] A, int[] B) {
            int m = A.length;
            int n = B.length;
            if (m > n) { // to ensure 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; // i is too small
                }
                else if (i > iMin && A[i-1] > B[j]) {
                    iMax = i - 1; // i is too big
                }
                else { // i is perfect
                    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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  • 原文地址:https://www.cnblogs.com/Yanss/p/12068687.html
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