我正在解决可汗学院二元搜索问题,第三步是要求对“...帮助可视化搜索需要多长时间”进行一些基本添加。
提示要求“...添加一个println()语句,显示找到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总数。您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println()带有该变量的计数器成功完成了此操作。运行代码完美无缺,但是,可汗学院的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);
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注意:我还尝试使用guessand添加返回数组guessTotal,然后从result. 那也满足了他们的要求,但也没有通过。
小智 5
这里正确答案
/* 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);
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