我正在处理 Khan Academy Binary Search problem第三步是要求对“...帮助可视化搜索需要多长时间”进行一些基本的补充。
提示要求“...添加一个 println()
语句,该语句显示查找结果所需的猜测总数。
您的函数应该只在找到目标时打印猜测的总数。你的函数不应该打印每个循环的猜测次数。”
我已经使用递增计数器和带有该变量的 println()
成功完成了这项工作。运行代码可以完美运行,但是可汗学院预建环境不会让我通过这一步。他们还有其他期待的方式吗?
代码在这里:
/* Returns either the index of the location in the array,
or -1 if the array did not contain the targetValue */
var doSearch = function(array, targetValue) {
var min = 0;
var max = array.length - 1;
var guess;
var guessTotal = 0;
while(min <= max){
guess = Math.floor((min + max) / 2);
println("You guessed " + guess);
if(array[guess] === targetValue){
println(guessTotal);
return guess;
}
else if (array[guess] < targetValue){
min = guess + 1;
}
else{
max = guess -1;
}
guessTotal ++;
}
return -1;
};
var primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];
var result = doSearch(primes, 73);
println("Found prime at index " + result);
Program.assertEqual(doSearch(primes, 73), 20);
注意:我还尝试使用 guess
和 guessTotal
添加返回数组,然后从 result
打印它们。这也符合他们的要求,但也没有通过。
最佳答案
正确答案
/* Returns either the index of the location in the array,
or -1 if the array did not contain the targetValue */
var doSearch = function(array, targetValue) {
var min = 0;
var max = array.length - 1;
var guess;
var guessTotal = 0;
while(min <= max) {
guess = Math.floor((max + min) / 2);
println(guess);
guessTotal++;
if (array[guess] === targetValue) {
println(guessTotal);
return guess;
}
else if (array[guess] < targetValue) {
min = guess + 1;
}
else {
max = guess - 1;
}
}
return -1;
};
var primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];
var result = doSearch(primes, 73);
println("Found prime at index " + result);
Program.assertEqual(doSearch(primes, 73), 20);
关于khan-academy - 可汗学院二分搜索算法第 3 步正确,但未通过。,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30382288/