【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