确定一个数字数组是否可以分成两个数组，每个数组包含相同的数字总和
下面是一个代码，用于确定一个数字数组是否可以分成两个数组，每个数组包含相同的数字总和.例如:{1, 3 ,2, 6} 可分为 {6} 和 {1,2,3}，因此返回 true而{1,5,7}不能一分为二，平衡数组，因此返回false
Below is a code to determine whether or not can an array of numbers can be divided into two arrays, with each array holding the same sum of numbers. for example: {1, 3 ,2, 6} can be divided into {6} and {1,2,3}, therefore return true while {1,5,7} can't be divided into two, balanced array, therefore return false
public boolean canBalance(int[] nums) {
for (int i = 0; i < nums.length; i++) {
int sum = 0;
for (int j = 0; j < i; j++) sum += nums[j];
for (int j = i; j < nums.length; j++) sum -= nums[j];
if (sum == 0) return true;
}
return false;
}
这是codingbat练习的公认答案，我特别不理解这篇文章:
it's an accepted answer for an exercise in codingbat, I don't understand this piece in particular:
for (int j = 0; j < i; j++) sum += nums[j];
for (int j = i; j < nums.length; j++) sum -= nums[j];
不进行迭代通常以 { 开始并以 } 结束?如果 sum == 0 意味着它可以平衡怎么办?我试过在一张纸上记下它，数组为 {1,3,2,6}，总和为 26，返回 false，很明显 {1,3,2,6} 应该返回 true.
doesn't for iteration usually starts with { and ends with } ? and how come if sum == 0 means it can be balanced? I have tried noting it down on piece of paper with array of {1,3,2,6} and had sum of 26, which returns false, where it's obvious that {1,3,2,6} should return true.
我想我看错了代码，但我不知道是哪个.或者可能算法是假的，但在codingbat中被接受了
I think I misread the code, I don't know which, though. Or maybe the algorithm is false, but it was accepted in codingbat
推荐答案
两个for循环用于对数组的两部分进行加权，找到数组的数组平衡点.
The two for-loops are for weighing the two parts of the array, to find the array balancing point of an array.
这样想:
你有一个空的天平，在外部 for 循环的第一次迭代中，i 为零.
You have a empty balance scale, in the first iteration of the outer for loop, i is zero.
第一个for循环，这里j为0，i为0 i j 为假，所以它不会进入第一个 for 循环，而是进入第二个 for 循环并从 sum 中减去所有数字.
It comes to first for loop, here j is 0 and i is 0 i < j is false, so it doesn't enter the first for-loop and it goes into the second for-loop and subtracts all the numbers from sum.
从外部 for 循环的第二次迭代开始，它开始进入第一个 for 循环并开始将数组的元素一一添加到总和中.
From second iteration onwards of the outer for-loop, it starts entering the first for-loop and starts adding the elements of the array one-by-one to the sum.
在图片中，就像从一个空的天平秤开始，将所有元素添加到第二个秤，然后将一个元素移动到第一个秤，像这样:
In pictures, it is like starting with an empty balance scale, adding all the elements into the second scale, and moving one by one element to the first scale, like this:
最后，如果和为零，则数组可以平衡，所以返回true.如果总和不为 0，则为不平衡.
In the end, if the sum is zero, then the array can be balanced, so return true. If the sum isn't 0, it's unbalanced.
中的值由循环平衡，如下所示:
The values in the are balanced by the loops like this:
i为0时外层for循环的迭代
循环 2 ->i(0) j(0) 减1，和为-1
循环 2 ->i(0) j(1) 减 3，和为 -4
循环 2 ->i(0) j(2) 减2，和为-6
循环 2 ->i(0) j(3) 减 6，和为 -12
Iteration of outer for loop when i is 0
Loop 2 -> i(0) j(0) Subtract 1, sum is -1
Loop 2 -> i(0) j(1) Subtract 3, sum is -4
Loop 2 -> i(0) j(2) Subtract 2, sum is -6
Loop 2 -> i(0) j(3) Subtract 6, sum is -12
i为1时外层for循环的迭代
循环 1 ->i(1) j(0) 加1，和为1
循环 2 ->i(1) j(1) 减3，和为-2
循环 2 ->i(1) j(2) 减2，和为-4
循环 2 ->i(1) j(3) 减 6，和为 -10
Iteration of outer for loop when i is 1
Loop 1 -> i(1) j(0) Add 1, sum is 1
Loop 2 -> i(1) j(1) Subtract 3, sum is -2
Loop 2 -> i(1) j(2) Subtract 2, sum is -4
Loop 2 -> i(1) j(3) Subtract 6, sum is -10
i为2时外层for循环的迭代
循环 1 ->i(2) j(0) 加1，和为1
循环 1 ->i(2) j(1) 加3，和为4
循环 2 ->i(2) j(2) 减2，和为2
循环 2 ->i(2) j(3) 减 6，和为 -4
Iteration of outer for loop when i is 2
Loop 1 -> i(2) j(0) Add 1, sum is 1
Loop 1 -> i(2) j(1) Add 3, sum is 4
Loop 2 -> i(2) j(2) Subtract 2, sum is 2
Loop 2 -> i(2) j(3) Subtract 6, sum is -4
i为3时外部for循环的迭代
循环 1 ->i(3) j(0) 加1，和为1
循环 1 ->i(3) j(1) 加3，和为4
循环 1 ->i(3) j(2) 加2，和为6
循环 2 ->i(3) j(3) 减 6，和为 0
Iteration of outer for loop when i is 3
Loop 1 -> i(3) j(0) Add 1, sum is 1
Loop 1 -> i(3) j(1) Add 3, sum is 4
Loop 1 -> i(3) j(2) Add 2, sum is 6
Loop 2 -> i(3) j(3) Subtract 6, sum is 0
最终结果为真，因此数组可以平衡
public class Test {
public static void main(String[] args) {
int[] test = { 1, 3, 2, 6 };
System.out.println("
Final result is "+canBalance(test));
}
public static boolean canBalance(int[] nums) {
for (int i = 0; i < nums.length; i++) {
System.out.println("
Iteration of outer for loop when i is " + i);
int sum = 0;
for (int j = 0; j < i; j++){
sum += nums[j];
System.out.println("Loop 1 -> i(" +i + ") j("+j + ") Add "+nums[j] + ", sum is "+sum+" ");
}
for (int j = i; j < nums.length; j++){
sum -= nums[j];
System.out.println("Loop 2 -> i(" +i + ") j("+j + ") Subtract "+nums[j] + ", sum is "+sum+" ");
}
if (sum == 0)
return true;
}
return false;
}
}
如果你想允许数组元素之间的混洗，你可以使用递归如下(注释是不言自明的)
public class Test {
public static void main(String[] args) {
int[] original = { 10, 2, 24, 32 };
System.out.println(canDivideArray(original));
}
private static boolean canDivideArray(int[] originalArray) {
int total = 0;
for (int number : originalArray) {
total += number;
}
// check if sum == 2x for any value of x
if (total % 2 != 0) {
return false;
} else {
// sum of each half array should be x
total /= 2;
}
return isTotal(originalArray, originalArray.length, total);
}
private static boolean isTotal(int array[], int n, int total) {
// successful termination condition
if (total == 0) {
return true;
}
// unsuccessful termination when elements have finished but total is not reached
if (n == 0 && total != 0){
return false;
}
// When last element is greater than total
if (array[n - 1] > total)
return isTotal(array, n - 1, total);
//check if total can be obtained excluding the last element or including the last element
return isTotal(array, n - 1, total - array[n - 1]) || isTotal(array, n - 1, total);
}
}
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