[ Javascript ] - 20140901-题目1

[ Javascript ] - 20140901-题目1

A,B有两个数组，A递增数组，B递减数组。
请用最优的js代码找出A、B两个数组中交集的第K个元素。
比如：
var A = [3,4,5,6,7,8,9];
var B = [12,10,8,6];
交集的元素有6,8,
第1个元素是6，第2个元素是8。
那么js是这样：
function getNum(a, b, k){
   // code
}
getNum(A, B, 1) -> 6,
getNum(A, B, 2) -> 8,


PS:
1. 回复时注意加上下面这句话，才会有语法高亮或格式缩进。
```javascript
// you code
```
2. 粘贴代码时请使用shift+tab，缩进前面的空白。  

var getNum = function(A,B,index){
var num = [];
for(var i =0;i< A.length;i++){
if(B.indexOf(A[i])!== -1){
num.push(A[i]);
B.length = B.indexOf(A[i]);
}
}
if(index < num.length){
return num[index];
}else{
return null;
}
};
getNum([1,2,3,4],[3,2,0],0);

var getNum = function(a, b, index) {
  var i = 0;
  var j = b.length - 1;
  var k = 0;
  while (i < a.length && j >= 0) {
    if (a[i] == b[j]) {
      k++;
      if (k == index) {
        return a[i];
      }
      i++;
      j--;
    } else if (a[i] > b[j]) {
      j--;
    } else {
      i++;
    }
  }
  return undefined;
};

var A = [3,4,5,6,7,8,9];
var B = [12,10,8,6];
console.info(getNum(A, B, 1));
console.info(getNum(A, B, 2));

function getNum(a,b,num){
    var obj = {},
          cont = a.concat(b),
          result = [];
    for(var i=0;i<cont.length;i++){
        obj[cont[i]]?result.push(cont[i]):obj[cont[i]] = {};
    };
    return (num?result[num-1]:result);
};
var A = [3,4,5,6,7,8,9];
var B = [12,10,8,6];
getNum(A, B, 1);
getNum(A, B, 2);

var getNum = function (a, b, k) {
    return a.filter(function (v) {return b.indexOf(v) > -1;})[k-1];
}
var A = [3,4,5,6,7,8,9];
var B = [12,10,8,6];
getNum(A, B, 1);
getNum(A, B, 2);

function getNum(a, b, k){
  var c =[];
  for(var i=0;i<a.length;i++){
    b.indexOf(a[i])>-1 && (c.push(a[i]));
  }
  return c[k-1];  
}
var A = [3,4,5,6,7,8,9];
var B = [12,10,8,6];
getNum(A, B, 1);
getNum(A, B, 2);

var a =  [3,4,5,6,7,8,9];
            var b = [12,10,8,6];
            function getNum( arr1, arr2, index ){
                var arr3 = [],
                    len = arr2.length,
                    temp = len;

                for(var i = 0 ; i < arr1.length ; i ++){
                    if(temp >= 0 ){
                        if(arr1[i] == arr2[temp-1]) {
                            arr3.push(arr2[temp-1]);
                            temp-=1;
                        }
                    }
                }
                return arr3[index-1]

            }

            console.log(getNum(a, b , 1));
            console.log(getNum(a, b , 2));

function getNum(a, b, k){
    return a.filter(function (v){ return b.indexOf(v)!=-1;})[k-1];
}