Median of Two Sorted Arrays

最新推荐文章于 2023-10-07 04:23:19 发布

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最新推荐文章于 2023-10-07 04:23:19 发布

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本文介绍了一个算法问题，即找到两个已排序数组的中位数，并提供了一个Java实现方案。该方法通过合并两个数组并利用分治思想来减少时间复杂度，最终达到O(log(m+n))的时间复杂度。

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    	int len1 = nums1.length;
    	int len2 = nums2.length;
    	if (len1 == 0 && len2%2==0) {
    		return (nums2[len2/2-1]+nums2[len2/2])/2.0;
    	}
    	if (len1 == 0 && len2%2==1) {
    		return nums2[len2/2];
    	}
    	if (len2 == 0 && len1%2==0) {
    		return (nums1[len1/2-1]+nums1[len1/2])/2.0;
    	}
    	if (len2 == 0 && len1%2==1) {
    		return nums1[len1/2];
    	}
    	int mid = (len1+len2)/2;
    	if ((len1+len2)%2 == 0) {
    	return (findKthNum(nums1, nums2, mid)+findKthNum(nums1, nums2, mid+1))/2.0;
    	} else {
    	return findKthNum(nums1, nums2, mid+1);
    	}
    }
    public static int findKthNum(int[] arr1, int[] arr2, int kth) {
		if (arr1 == null || arr2 == null) {
			throw new RuntimeException("Your arr is invalid!");
		}
		if (kth < 1 || kth > arr1.length + arr2.length) {
			throw new RuntimeException("K is invalid!");
		}
		int[] longs = arr1.length >= arr2.length ? arr1 : arr2;
		int[] shorts = arr1.length < arr2.length ? arr1 : arr2;
		int l = longs.length;
		int s = shorts.length;
		if (kth <= s) {
			return getUpMedian(shorts, 0, kth - 1, longs, 0, kth - 1);
		}
		if (kth > l) {
			if (shorts[kth - l - 1] >= longs[l - 1]) {
				return shorts[kth - l - 1];
			}
			if (longs[kth - s - 1] >= shorts[s - 1]) {
				return longs[kth - s - 1];
			}
			return getUpMedian(shorts, kth - l, s - 1, longs, kth - s, l - 1);
		}
		if (longs[kth - s - 1] >= shorts[s - 1]) {
			return longs[kth - s - 1];
		}
		return getUpMedian(shorts, 0, s - 1, longs, kth - s, kth - 1);
	}

	public static int getUpMedian(int[] a1, int s1, int e1, int[] a2, int s2, int e2) {
		int mid1 = 0;
		int mid2 = 0;
		int offset = 0;
		while (s1 < e1) {
			mid1 = (s1 + e1) / 2;
			mid2 = (s2 + e2) / 2;
			offset = ((e1 - s1 + 1) & 1) ^ 1;
			if (a1[mid1] > a2[mid2]) {
				e1 = mid1;
				s2 = mid2 + offset;
			} else if (a1[mid1] < a2[mid2]) {
				s1 = mid1 + offset;
				e2 = mid2;
			} else {
				return a1[mid1];
			}
		}
		return Math.min(a1[s1], a2[s2]);
	}
}