1. two Sum

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].

 

 

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
     var hash = {},len = nums.length;
    
    //其实这里是有重复的值可能性的，比如[3,3],那第二个会覆盖第一个
    for(var i = 0;i < len;i++){
        
        hash[nums[i]] = i;
    }
    
    
    //由于题目说过有且只有一组解，所以上面覆盖也没关系，因为我们只找后面一个可行解。
    for(var j = 0;j < len;j++){
        if(hash[target - nums[j]]){
            return [j,hash[target - nums[j]]];
        }
    }
};

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
  
    //双指针，思路就是先排序，两边向中间靠拢
    var old = nums.slice(),
        len = old.length,
        i = 0,
        j = len -1;
    nums.sort();
    while(i<len&&j > -1){
        
        var temp = nums[i] + nums[j];
        if(temp === target){
            var fir = old.indexOf(nums[i]);
            //这里要防止重复[3,3]求6
            
            var sec = old.lastIndexOf(nums[j])
            return [fir,sec];
        }
        
        if(temp < target){
            i++;
        }else{
            j--;
        }
    }
    
};