1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解法一

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */

var twoSum = function(nums, target) {
    if ( !Array.isArray(nums) || Object.prototype.toString.call(target) !== "[object Number]" ) {
         return;
    }  
var i,j,len = nums.length;

  for ( i = 0; i < len; i ++ ){
    for ( j = i + 1; j < len; j ++ ){
      if ( nums[i] + nums[j] === target ) 
          return [i, j];
    }
  }
};

解法二

var twoSum = function(nums, target) {
    if ( !Array.isArray(nums) || Object.prototype.toString.call(target) !== "[object Number]" ) 
        return;

  var arr = [],i,j,len = nums.length;
  //构造哈希表
  for ( i = 0; i < len; i ++ ){
    arr[nums[i]] = i;
  }

  for ( i = 0; i < len; i ++ ){
    j = arr[ target - nums[i] ];
    // 例如 4+4=8，但只有一个4
    if ( j !== undefined && i !== j ) return [ i, j ];
  }
};