Comments (1)
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var findMedianSortedArrays = function(nums1, nums2) {
let m = nums1.length
let n = nums2.length
let k1 = Math.floor((m + n + 1) / 2)
let k2 = Math.floor((m + n + 2) / 2)
return (findMedianSortedArraysCore(nums1, 0, nums2, 0, k1) + findMedianSortedArraysCore(nums1, 0, nums2, 0, k2)) / 2
};
/**
*
* @param {number[]} nums1
* @param {number[]} nums2
* @param {number} i
* @param {number} j
* @param {number} k
* @return {number}
*/
const findMedianSortedArraysCore = (nums1, i, nums2, j, k) => {
// 如果数组起始位置已经大于数组长度-1
// 说明已经是个空数组
// 直接从另外一个数组里取第k个数即可
if (i > nums1.length - 1) {
return nums2[j + k - 1]
}
if (j > nums2.length - 1) {
return nums1[i + k - 1]
}
// 如果k为1
// 就是取两个数组的起始值里的最小值
if (k === 1) {
return Math.min(nums1[i], nums2[j])
}
// 取k2为(k/2)或者数组1的长度或者数组2的长度的最小值
// 这一步可以避免k2大于某个数组的长度(长度为从起始坐标到结尾)
let k2 = Math.floor(k / 2)
let length1 = nums1.length - i
let length2 = nums2.length - j
k2 = Math.min(k2, length1, length2)
let value1 = nums1[i + k2 - 1]
let value2 = nums2[j + k2 - 1]
// 比较两个数组的起始坐标的值
// 如果value1小于value2
// 就舍弃nums1前i + k2部分
// 否则舍弃nums2前j + k2部分
if (value1 < value2) {
return findMedianSortedArraysCore(nums1, i + k2, nums2, j, k - k2)
} else {
return findMedianSortedArraysCore(nums1, i, nums2, j + k2, k - k2)
}
}
from interview.
Related Issues (20)
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- Day397:手写 Vue.extend 实现?
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- Day402: 实现一个render(str,parameter)方法,将str中的占位符用parameter填充?(腾讯) HOT 1
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from interview.