可汗学院算法:二分搜索解决方案
Khan academy Algorithm: Binary Search Solution
我在可汗学院研究算法:
https://www.khanacademy.org/computing/computer-science/algorithms/binary-search/p/challenge-binary-search
下面的大部分代码结果为 -1 ?这是为什么?
所以二分搜索不会高效工作?
var doSearch = function(array, targetValue) {
var min = 0;
var max = array.length - 1;
var guess;
while(min < max) {
guess = Math.floor((max + min) / 2);
if (array[guess] === targetValue) {
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];
for(var i=0;i<primes.length;i++){
var result = doSearch(primes,primes[i]);
console.log("Found prime at index of " + primes[i] +" @ " + result);
}
结果:
Found prime at index of 2 @ 0
Found prime at index of 3 @ -1
Found prime at index of 5 @ 2
Found prime at index of 7 @ 3
Found prime at index of 11 @ -1
Found prime at index of 13 @ 5
Found prime at index of 17 @ 6
Found prime at index of 19 @ -1
Found prime at index of 23 @ 8
Found prime at index of 29 @ -1
Found prime at index of 31 @ 10
Found prime at index of 37 @ -1
Found prime at index of 41 @ 12
Found prime at index of 43 @ 13
Found prime at index of 47 @ -1
Found prime at index of 53 @ 15
Found prime at index of 59 @ 16
Found prime at index of 61 @ -1
Found prime at index of 67 @ 18
Found prime at index of 71 @ 19
Found prime at index of 73 @ -1
Found prime at index of 79 @ 21
Found prime at index of 83 @ -1
Found prime at index of 89 @ 23
Found prime at index of 97 @ -1
我错过了什么?
您过早终止循环 - min == max
是一个有效条件。
将循环更改为
while(min <= max) {
guess = Math.floor((max + min) / 2);
if (array[guess] === targetValue) {
return guess;
}
else if (array[guess] < targetValue) {
min = guess + 1;
}
else {
max = guess - 1;
}
}
我得到
的输出
VM153:30 Found prime at index of 2 @ 0
VM153:30 Found prime at index of 3 @ 1
VM153:30 Found prime at index of 5 @ 2
VM153:30 Found prime at index of 7 @ 3
VM153:30 Found prime at index of 11 @ 4
VM153:30 Found prime at index of 13 @ 5
VM153:30 Found prime at index of 17 @ 6
VM153:30 Found prime at index of 19 @ 7
VM153:30 Found prime at index of 23 @ 8
VM153:30 Found prime at index of 29 @ 9
VM153:30 Found prime at index of 31 @ 10
VM153:30 Found prime at index of 37 @ 11
VM153:30 Found prime at index of 41 @ 12
VM153:30 Found prime at index of 43 @ 13
VM153:30 Found prime at index of 47 @ 14
VM153:30 Found prime at index of 53 @ 15
VM153:30 Found prime at index of 59 @ 16
VM153:30 Found prime at index of 61 @ 17
VM153:30 Found prime at index of 67 @ 18
VM153:30 Found prime at index of 71 @ 19
VM153:30 Found prime at index of 73 @ 20
VM153:30 Found prime at index of 79 @ 21
VM153:30 Found prime at index of 83 @ 22
VM153:30 Found prime at index of 89 @ 23
VM153:30 Found prime at index of 97 @ 24
假设您的列表只有 2 个质数,并且您搜索更大的一个。您的循环将针对较小的值进行测试,失败,并将 min
设置为与 max
相同,因此循环将终止 ,然后再检查该值 。
你的循环守卫应该是min <= max
。
我在可汗学院研究算法: https://www.khanacademy.org/computing/computer-science/algorithms/binary-search/p/challenge-binary-search 下面的大部分代码结果为 -1 ?这是为什么? 所以二分搜索不会高效工作?
var doSearch = function(array, targetValue) {
var min = 0;
var max = array.length - 1;
var guess;
while(min < max) {
guess = Math.floor((max + min) / 2);
if (array[guess] === targetValue) {
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];
for(var i=0;i<primes.length;i++){
var result = doSearch(primes,primes[i]);
console.log("Found prime at index of " + primes[i] +" @ " + result);
}
结果:
Found prime at index of 2 @ 0
Found prime at index of 3 @ -1
Found prime at index of 5 @ 2
Found prime at index of 7 @ 3
Found prime at index of 11 @ -1
Found prime at index of 13 @ 5
Found prime at index of 17 @ 6
Found prime at index of 19 @ -1
Found prime at index of 23 @ 8
Found prime at index of 29 @ -1
Found prime at index of 31 @ 10
Found prime at index of 37 @ -1
Found prime at index of 41 @ 12
Found prime at index of 43 @ 13
Found prime at index of 47 @ -1
Found prime at index of 53 @ 15
Found prime at index of 59 @ 16
Found prime at index of 61 @ -1
Found prime at index of 67 @ 18
Found prime at index of 71 @ 19
Found prime at index of 73 @ -1
Found prime at index of 79 @ 21
Found prime at index of 83 @ -1
Found prime at index of 89 @ 23
Found prime at index of 97 @ -1
我错过了什么?
您过早终止循环 - min == max
是一个有效条件。
将循环更改为
while(min <= max) {
guess = Math.floor((max + min) / 2);
if (array[guess] === targetValue) {
return guess;
}
else if (array[guess] < targetValue) {
min = guess + 1;
}
else {
max = guess - 1;
}
}
我得到
的输出VM153:30 Found prime at index of 2 @ 0
VM153:30 Found prime at index of 3 @ 1
VM153:30 Found prime at index of 5 @ 2
VM153:30 Found prime at index of 7 @ 3
VM153:30 Found prime at index of 11 @ 4
VM153:30 Found prime at index of 13 @ 5
VM153:30 Found prime at index of 17 @ 6
VM153:30 Found prime at index of 19 @ 7
VM153:30 Found prime at index of 23 @ 8
VM153:30 Found prime at index of 29 @ 9
VM153:30 Found prime at index of 31 @ 10
VM153:30 Found prime at index of 37 @ 11
VM153:30 Found prime at index of 41 @ 12
VM153:30 Found prime at index of 43 @ 13
VM153:30 Found prime at index of 47 @ 14
VM153:30 Found prime at index of 53 @ 15
VM153:30 Found prime at index of 59 @ 16
VM153:30 Found prime at index of 61 @ 17
VM153:30 Found prime at index of 67 @ 18
VM153:30 Found prime at index of 71 @ 19
VM153:30 Found prime at index of 73 @ 20
VM153:30 Found prime at index of 79 @ 21
VM153:30 Found prime at index of 83 @ 22
VM153:30 Found prime at index of 89 @ 23
VM153:30 Found prime at index of 97 @ 24
假设您的列表只有 2 个质数,并且您搜索更大的一个。您的循环将针对较小的值进行测试,失败,并将 min
设置为与 max
相同,因此循环将终止 ,然后再检查该值 。
你的循环守卫应该是min <= max
。