我正在尝试编写一个简单的二进制搜索，但我无法让它返回位置，它找到但不会返回它

Question

我正在尝试编写一个简单的二进制搜索函数来查找数组中的值并返回它的位置，如果不存在则返回 -1。通过控制台日志，我可以看到它在内部工作，但没有返回我想要的。它总是返回 undefined...

function binarySearch(arr, q) {
    // set left right middle
    let left = 0,
        right = arr.length - 1,
        middle = Math.floor(arr.length / 2);
    // check if the middle is q
    if (arr[middle] === q) {
        return middle;
    }
    console.log(checkMiddle(left, right, middle));

    let answer = checkMiddle(left, right, middle);

    function checkMiddle(l, r, m) {
        console.log(r - l);
        if (r - l === 1) {
            console.log('not found');
            return -1;
        }
        if (arr[m] === q) {
            // if yes return middle
            console.log('found');
            console.log(m);
            return m;
        }

        //if not find if q is > or < middle
        if (q > arr[m]) {
            // if > move left to middle
            l = m;

            //move middle to new middle
        } else {
            //if < move right to middle
            r = m;
        }
        m = Math.floor((r + l) / 2);
        //search again
        console.log(l, r, m);

        checkMiddle(l, r, m);
    }

    return answer;
}

let arr = [1, 2, 3, 4, 5, 6, 7, 55, 192, 5000, 666666, 8888888, 1055556565456],
    q = 10;
console.log(binarySearch(arr, q));

Answer 1

它返回 undefined 因为您没有从函数中返回任何内容，也没有将答案存储在变量中。

在递归函数中使用 return 只会返回到前一个调用堆栈。

定义一个全局变量，让你的递归函数在每次调用时将答案存储在其中，这样你就可以在函数完成后返回它。

下面的修改代码。

function binarySearch(arr, q) {
    // set left right middle
    let left = 0,
        right = arr.length - 1,
        middle = Math.floor(arr.length / 2);
    // check if the middle is q
    if (arr[middle] === q) {
        return middle;
    }
    //console.log(checkMiddle(left, right, middle));

    let answer = 0; // global variable which is modified by the checkMiddle function
    checkMiddle(left, right, middle);
    
    function checkMiddle(l, r, m) {
        console.log(r - l);
        if (r - l === 1) {
            console.log('not found');
            answer = -1;
            return;
        }
        if (arr[m] === q) {
            // if yes return middle
            console.log('found');
            console.log(m);
            answer = m;
            return;
        }

        //if not find if q is > or < middle
        if (q > arr[m]) {
            // if > move left to middle
            l = m;

            //move middle to new middle
        } else {
            //if < move right to middle
            r = m;
        }
        m = Math.floor((r + l) / 2);
        //search again
        console.log(l, r, m);

        checkMiddle(l, r, m);
    }
    console.log("answer : "+answer);
    return answer;
}

let arr = [1, 2, 3, 4, 5, 6, 7, 55, 192, 5000, 666666, 8888888, 1055556565456],
    q = 10;
console.log(binarySearch(arr, q));

Answer 2

我认为这是一种更简单的方法。

    let arr = [1, 2, 3, 4, 5, 6, 7, 55, 192, 5000, 666666, 8888888, 1055556565456];


    function binarySearch(arr, k, s, e){
        
      s = s || 0;
      e = e || arr.length;
      
      m = s + Math.floor((e - s) / 2);
        
        if (arr[m] === k)
            return m;
        
        if (k > arr[m]){
            return (m+1 < e)?binarySearch(arr, k, m+1, e):-1;
        }else if (m-1 >= s){
            return binarySearch(arr, k, s, m-1);
        }
        
        return -1;
    }


    console.log('answer: '+binarySearch(arr, 3));   //  2
    console.log('answer: '+binarySearch(arr, 10));  // -1
    console.log('answer: '+binarySearch(arr, 55));  //  7

这是一个尝试在迭代的每个步骤中查找“40”的示例。 （其中 V 是被检查数组中的值）

S = 0, E = 13, V = 7  (40 is higher than 7)
Search upper half...
S = 7, E = 13, V = 666666 (40 is lower than 666666)
Search lower half...
S = 7, E = 9, V = 192 (40 is lower than 192)
Search lower half...
S = 7, E = 7, V = 55 (40 is lower than 55)
Not found.