LeetCode 算法题解 js 版 (001 Two Sum)

LeetCode 算法题解 js 版 (001 Two Sum)

两数之和

https://leetcode.com/problems/two-sum/submissions/

https://leetcode-cn.com/problems/two-sum/submissions/

1. 暴力解法

Time complexity: O(n**2)

Space complexity: O(n)

"use strict";

/**
 * @author xgqfrms
 * @description leetcode problems: two-sum & https://leetcode.com/problems/two-sum/
 * @language JavaScript & ES6
 *
 * @param {Int Number Array} nums
 * @param {Int Number} target
 * @return {Int Number Array} result
 *
 */

let twoSum = (nums = [2, 7, 11, 15], target = 9) => {
    let result = [];
    for(let i =  0; i < nums.length; i++){
        for(let ii =  0; ii < nums.length; ii++){
            let temp_a = nums[i];
            let temp_b = nums[ii];
            if(ii !== i){
                let temp_result = temp_a + temp_b;
                if(temp_result === target){
                    // 1. reset;
                    // result = [];
                    // result.push(i);
                    // result.push(ii);
                    // 2. remove duplication
                    if(!result.includes(i)){
                       result.push(i);
                    }
                    if(!result.includes(ii)){
                       result.push(ii);
                    }
                    result.sort();
                }
            }else{
                // break;
            }
        }
    }
    return result;
};

2. 一遍 Hash Table

Time complexity: O(n)

Space complexity: O(n)

// Time complexity: O(n)
// Space complexity: O(n)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
  let result = [];
  // 空间换时间  HashMap / Object
  const obj = {};
  for(let i = 0; i < nums.length; i++) {
    let temp = target - nums[i];
    if(obj[nums[i]] !== undefined) {
      result = [obj[nums[i]], i];
      break;
    } else {
      obj[temp] = i;
    }
  }
 return result;
};

ES6

Map & WeakMap

bugs

// bug

/*

var twoSum = function(nums, target) {
    let result = [];
    for(let i = 0; i < target; i++) {
      let temp = target - i;
      if(nums.includes(temp) && nums.includes(i)) {
        result = [nums.indexOf(i), nums.indexOf(temp)];
        break;
      }
    }
   return result;
};

*/

best solutions

https://leetcode-cn.com/submissions/detail/112453419/

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    var map=new Map();
    for(let i=0;i<nums.length;i++){
        if(!map.has(target-nums[i])){
            map.set(nums[i],i);
        }else{
            return [map.get(target-nums[i]),i]
        }
    }
};
/**
 * 注：此题使用ES6的Map，使得时间复杂度降为O(n)
 * 关于Map和Set的使用要重点掌握！！！
 * 我觉得思路很棒，继续理解！
 */

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let i = nums.length;
    while(i > 1) {
        let last = nums.pop();
        if (nums.indexOf(target - last) > -1){
            return [nums.indexOf(target - last),nums.length];
        }
        i--
    }
}

refs

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