leetcode：4. Median of Two Sorted Arrays

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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);    }};

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