【LeetCode】4.Median of Two Sorted Arrays（Hard）解题报告

【LeetCode】4.Median of Two Sorted Arrays（Hard）解题报告

题目地址：https://leetcode.com/problems/median-of-two-sorted-arrays/description/
题目描述：

  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)).

Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

  此题非常重要，多看几遍。看到时间复杂度log想到二分法。

Solution:

//超级重要的一道题
//time : O(log(min(m,n)))
//space : O(1)
/*
参考博客：http://blog.csdn.net/chen_xinjia/article/details/69258706
index : 0   1   2    3      4     5
           L1   R1
num1 :  3   5 | 8    9                   4  cut1:左边有几个元素
num2 :  1   2   7  |  10    11    12      6  cut2:右边有几个元素
                L2    R2
num3 :  1   2   3    5    7  |  8    9    10    11    12
num3 -> num1+num2 -> num1  转换到只求一个

L1 <= R2
L2 <= R1

L1>R2 cut1 << (cutL , cut1-1)
L2>R1 cut2 >> (cut1+1 , cutR)
*/
class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if(nums1.length > nums2.length){
            return findMedianSortedArrays(nums2 , nums1);
        }
        int len = nums1.length + nums2.length;
        int cut1 = 0;
        int cut2 = 0;
        int cutL = 0;
        int cutR = nums1.length;
        while(cut1<=nums1.length){
            cut1 = (cutR - cutL)/2 + cutL;
            cut2 = len/2 - cut1;
            double L1 = (cut1 == 0) ? Integer.MIN_VALUE : nums1[cut1-1];
            double L2 = (cut2 == 0) ? Integer.MIN_VALUE : nums2[cut2-1];
            double R1 = (cut1 == nums1.length) ? Integer.MAX_VALUE : nums1[cut1];
            double R2 = (cut2 == nums2.length) ? Integer.MAX_VALUE : nums2[cut2];
            if(L1 > R2){
                cutR = cut1 - 1;
            }else if(L2 > R1){
                cutL = cut1 + 1;
            }else{
                if(len%2 == 0){
                    L1 = L1>L2?L1:L2;
                    R1 = R1return (L1+R1)/2;
                }else{
                    R1 = (R1 < R2) ? R1 : R2;
                    return R1;
                }
            }
        }
        return -1;
    }
}

Date：2018年2月22日