某个数组，通过交换使所有奇数都在前半所有偶数都在后半，复杂度O(N)。

function swaping3(A, i, j){
    if((i === j) || (i === j + 1)){
//　　if(i >= j){
        return;
    }

    //debugger;
    if(A[i]%2 === 1){
        if(A[j]%2 === 1){
            swaping3(A, ++i, j);
        }else{
            swaping3(A, ++i, --j);
        }
    }else{
        if(A[j]%2 === 1){
            var temp;
            temp = A[i];
            A[i] = A[j];
            A[j] = temp;
            swaping3(A, ++i, --j);
        }else{
            swaping3(A, i, --j);
        }
    }
//    return A;
}


function ARR(num, max){
    if(arguments.length < 2){
        alert("Not enough parameter ~");
        return;
    }

    var AA = [];
    for(var i = 0; i<num; i++){
        AA.push(    Math.ceil(Math.random() * max)    );
    }

    return AA;

//    var i = 0, j = num - 1;
//    swaping(this.AA, i, j);
//    return("Swap finished ~")
}

var OO = new Object();
OO.A = ARR(30, 100);
swaping3.call(OO, OO.A, 0, OO.A.length - 1);

上面是代码，十分简单，下面是执行结果：

OO.A = ARR(30, 100);
[85, 70, 44, 100, 37, 81, 97, 89, 67, 7, 73, 5, 83, 41, 33, 96, 80, 93, 73, 59, 89, 42, 27, 15, 49, 26, 81, 10, 62, 95]
swaping3.call(OO, OO.A, 0, 29);
undefined
OO.A
[85, 95, 81, 49, 37, 81, 97, 89, 67, 7, 73, 5, 83, 41, 33, 15, 27, 93, 73, 59, 89, 42, 80, 96, 100, 26, 44, 10, 62, 70]
OO.A = ARR(50, 10000);
[2124, 5000, 5082, 5998, 7476, 962, 5368, 9555, 3593, 5932, 2321, 8518, 2118, 186, 1496, 5914, 1605, 4569, 162, 7487, 4689, 2445, 3122, 8395, 8659, 2368, 1123, 2680, 673, 5098, 3679, 858, 9779, 50, 6818, 823, 8880, 8461, 1496, 942, 2316, 6356, 4691, 2397, 3756, 9451, 3103, 9179, 3108, 9892]
swaping3.call(OO, OO.A, 0, 49);
undefined
OO.A
[9179, 3103, 9451, 2397, 4691, 8461, 823, 9555, 3593, 9779, 2321, 3679, 673, 1123, 8659, 8395, 1605, 4569, 2445, 7487, 4689, 162, 3122, 5914, 1496, 2368, 186, 2680, 2118, 5098, 8518, 858, 5932, 50, 6818, 5368, 8880, 962, 1496, 942, 2316, 6356, 7476, 5998, 3756, 5082, 5000, 2124, 3108, 9892]

功能正常～复杂度也满足O(N)～