leetcode：4. Median of Two Sorted Arrays

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

主要思路是每次排除1/2k个

/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var findMedianSortedArrays = function(nums1, nums2) {
if((nums1.length + nums2.length) % 2 == 1) {
return findKthNumber(nums1,0,nums2,0,Math.ceil((nums1.length + nums2.length) / 2));
} else {
var mid1 = findKthNumber(nums1,0,nums2,0,(nums1.length + nums2.length) / 2);
var mid2 = findKthNumber(nums1,0,nums2,0,(nums1.length + nums2.length) / 2 + 1);
return (mid1 + mid2)/2;
}
};

var findKthNumber = function(nums1, begin1, nums2, begin2, k) {
var length1 = nums1.length - begin1;
var length2 = nums2.length - begin2;
// 保证nums1比nums2短
if(length1 > length2) {
return findKthNumber(nums2, begin2, nums1, begin1, k);
}
// 边界条件：第一个元素
if(k === 1) {
return nums1[begin1] < nums2[begin2] ? nums1[begin1] : nums2[begin2];
}
// 边界条件：一条数列已空
if(length1 === 0) {
return nums2[begin2 + k - 1];
}
// 如果两者都比Math.floor(k/2)长
var index = length1 < Math.floor(k/2) ? length1 : Math.floor(k/2);
// 每次排除k/2
if(nums1[begin1 + index - 1] < nums2[begin2 + index - 1]) {
return findKthNumber(nums1, begin1 + index, nums2, begin2, k - index);
} else {
return findKthNumber(nums1, begin1, nums2, begin2 + index, k - index);
}
};