每日一Leet

成日都冇心機操 開個po玩吓

const numberOfSteps = function(num) {
    let totalStep = 0
    const runStep = (num) => {
        if(num === 0) return 
        totalStep = totalStep + 1
        const isEven = num % 2 === 0
        return runStep(isEven ? num / 2 : num -1)
    }
    runStep(num)
    return totalStep
};

大家隨便改正

bitcount()

咁多人leetcode

諗真啲

const step = (num) => {
let bits = 0;
while (num > 0) {
num == num >> 1;
bits += 1;
}
return bits;
}

sor 眼殘冇睇題目唔駛理

學到嘢

const step = (num) => {
let bits = 0;
while (num > 0) {
bits += num & 1 + 1;
num == num >> 1;
bits += 1;
}
return bits - 1;
}

num & 1 係masking，apply bitwise and 去攞least significant bit。
E.g. 5 & 1 = 101 & 001 = 1，可以用黎判斷單雙數。
num >> 1 係right shift，將bits向右移一格。
E g. 5 >> 1 = 101 >> 1 = 10 = 2，個result同math.floor(num /2)等價。

class Solution {
public:
    int numberOfSteps(int num) {
        int step=0;
        while(num!=0)
        {
            if(num%2==0)
                num/=2;
            else
                num-=1;
            step+=1;
        }
        return step;
    }
};

LM學嘢

突然醒起recursion可以寫得clean啲

const numberOfSteps = function(num, totalSteps = 0) {
        if(num === 0) return totalSteps
        totalSteps = totalSteps + 1

        return numberOfSteps(num % 2 === 0 ? num / 2 : num -1, totalSteps)
};

但呢條嚟講唔係好複雜好似咁寫好啲

special case: num = 0

抄po狗

走啦 js撚
唔好獻世唔該

Two Sum
Given an array of integer nums and an integer target, return indices of the two numbers such that they add up to the target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.
Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

- 2 <= nums.length <= 104
- -109 <= nums[i] <= 109
- -109 <= target <= 109
- Only one valid answer exists. 

Answer:

const twoSum = (nums, target, numMap={}, step=0)=> {
    if (target-nums[step] in numMap) {
            return [numMap[target-nums[step]], step]
    }
    
    numMap[nums[step]] = step
    return twoSum(nums, target, numMap, step+1)
}; 

用js唔理memory usage