求审稿机会：人体工程学，腰肌劳损，腰痛，脊柱力学等# Immigration - 落地生根
z*o
1 楼
开始刷题，提交之后告诉Runtime: 116 ms，如何知道这个运行时间好坏？能否看到别
人的提交？看到有Java的Runtime 4ms，差很多，自己的算法好坏如何判断？
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Note:
Each element in the result should appear as many times as it shows in both
arrays.
The result can be in any order.
难度Easy。
我的代码：
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersect = function(nums1, nums2) {
if ( nums1.length === 0 || nums2.length === 0) {
return [];
}
// compare numbers
var compare = function(num1, num2){
return parseInt(num1) - parseInt(num2);
}
// sort
nums1 = nums1.sort(compare);
nums2 = nums2.sort(compare);
// get intersaction
return intersaction(nums1, nums2);
};
function intersaction(nums1, nums2){
var p1 = 0;
var p2 = 0;
var result = [];
while (p1 < nums1.length && p2 < nums2.length) {
if (nums1[p1] === nums2[p2]) {
result.push(nums1[p1]);
p1++;
p2++;
} else if (nums1[p1] < nums2[p2]) {
p1++;
} else {
p2++;
}
}
return result;
}
人的提交？看到有Java的Runtime 4ms，差很多，自己的算法好坏如何判断？
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Note:
Each element in the result should appear as many times as it shows in both
arrays.
The result can be in any order.
难度Easy。
我的代码：
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersect = function(nums1, nums2) {
if ( nums1.length === 0 || nums2.length === 0) {
return [];
}
// compare numbers
var compare = function(num1, num2){
return parseInt(num1) - parseInt(num2);
}
// sort
nums1 = nums1.sort(compare);
nums2 = nums2.sort(compare);
// get intersaction
return intersaction(nums1, nums2);
};
function intersaction(nums1, nums2){
var p1 = 0;
var p2 = 0;
var result = [];
while (p1 < nums1.length && p2 < nums2.length) {
if (nums1[p1] === nums2[p2]) {
result.push(nums1[p1]);
p1++;
p2++;
} else if (nums1[p1] < nums2[p2]) {
p1++;
} else {
p2++;
}
}
return result;
}
