在Javascript中过滤数组以找到3个连续数字

Question

给定一个未排序的正整数数组，编写一个函数，查找3个连续数字（升序或降序）的运行，并返回此类运行开始的索引。如果未找到此类运行，则返回null。

function findConsecutiveRuns(input:Array):Array

示例：[1,2,3,5,10,9,8,9,10,11,7]将返回[0,4,6,7]

我的JS技能有点生疏，这是我对此的尝试......

var numArray = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var newNumArray = []; 

for(var i = 1; i < numArray.length; i++) {
    if ((numArray[i] - numArray[i-1] != 1) || (numArray[i] + numArray[i+1] !=1)  {
        return 0;
    }
    else {
    newNumArray.push(numArray[i]);
    }
}
alert(newNumArray);

Answer 1

下面：

function f ( arr ) {
    var diff1, diff2, result = [];

    for ( var i = 0, len = arr.length; i < len - 2; i += 1 ) {
        diff1 = arr[i] - arr[i+1];
        diff2 = arr[i+1] - arr[i+2];
        if ( Math.abs( diff1 ) === 1 && diff1 === diff2 ) {
            result.push( i );
        }        
    }

    return result.length > 0 ? result : null;
}

现场演示： http://jsfiddle.net/Cc4DT/1/