4.寻找两个正序数组的中位数 about leetcode

[zzcyes]

题解

方法一: 函数

利用sort()排序,排序后直接取中位数

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
    let mergeArray = [...nums1,...nums2].sort((a, b) => a - b);
    return medianNums(mergeArray);
};
var medianNums = function(arr){
    let length = arr.length
    if(length%2==0){
        return (arr[length/2]+arr[length/2-1])/2
    }else{
        return arr[(length-1)/2]
    }
}

方法二: 二分查找

来自官方的图

来自lucifer

/**
 * 二分解法
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
  // make sure to do binary search for shorten array
  if (nums1.length > nums2.length) {
    [nums1, nums2] = [nums2, nums1]
  }
  const m = nums1.length
  const n = nums2.length
  let low = 0
  let high = m
  while(low <= high) {
    const i = low + Math.floor((high - low) / 2)
    const j = Math.floor((m + n + 1) / 2) - i

    const maxLeftA = i === 0 ? -Infinity : nums1[i-1]
    const minRightA = i === m ? Infinity : nums1[i]
    const maxLeftB = j === 0 ? -Infinity : nums2[j-1]
    const minRightB = j === n ? Infinity : nums2[j]

    if (maxLeftA <= minRightB && minRightA >= maxLeftB) {
      return (m + n) % 2 === 1
        ? Math.max(maxLeftA, maxLeftB)
        : (Math.max(maxLeftA, maxLeftB) + Math.min(minRightA, minRightB)) / 2
    } else if (maxLeftA > minRightB) {
      high = i - 1
    } else {
      low = low + 1
    }
  }
};