算法大挑战

[fengjian_428]

请编写以下函数 int MajorityElement(int array[],int n);
该函数返回数组array中的多数元素。多数元素是指在占绝对多数（至少51%）的一个值。如果多数元素不存在，那么返回常量NoMajorityElement,该函数必须满足下面的条件：
1. 必须以O(N)时间运行。
2. 必须使用O（1）的附加空间。换句话说，可用个别的临时变量，而不可以使用任何的临时数组。并且不能使用递归解决，这是因为随着递归层数加深，会需要空间来存储栈帧。
3. 不能改变数组中的任何元素的值。

大家一起讨论下啦

[鸿悯]

晕，这不是经典的主元素问题么，？

[qlzf11140820]

关注

[lfywy]

学习

[Steve]

[Quote=引用 86 楼 fengjian_428 的回复:]
12L的已经是正确的了， 对于题目要求：抓出大于50%
[/Quote]
嗯,>50%的话,52的反例也不成立.
12楼正确.

[btsy2000]

注册一下~

[suiyili]

Sorry my PC doesn't have chinese spell.

If an array N has S elements, then 51% means that there must be S*0.51 number of same elements.

Here, to be simple, suppose we have 100 elements, then ramdomly find an element, ok, we use the first element N[0], then, compared with N[0], go through every item in the array, after that we can get how many elements which are more than N[0] (part 1), and how many are equal to it (part 2), how many are less than it(part3). When iterating the array, we randomly pick up N[p1] which is in part1 (N[p1] < N[0]) and N[p2] in part2 (N[0] < N[p2]).

if no number in the 3 part is more then 51, return false.

otherwise, if part2 >= 51, ok N[0] is the one.
if either part1 or part3 is more then 51, here suppose part3 is more than 51. since N[p2] in part2( N[0] < N[p2]), we can use both of values N[0] and N[p2] to go through the array again. this time, just focus on how many are between N[0] and N[p2], how many are equal to N[p2] and how many are more then N[p2].

Now we have new three parts (par1, between N[0] and N[p2], part2 equal to N[p2], part3, more than N[p2]).
at the same time, don't forget to pick up a value in part1 and part3 as new N[p1] and new N[p2] for next time compare.
continue above process, until find or fail.

[suiyili]

there is some mistake above, here is the correct one.

If an array N has S elements, then 51% means that there must be S*0.51 number of same elements.

Here, to be simple, suppose we have 100 elements, then ramdomly find an element, ok, we use the first element N[0], then, compared with N[0], go through every item in the array, after that we can get how many elements which are more than N[0] (part 1), and how many are equal to it (part 2), how many are less than it(part3). When iterating the array, we randomly pick up N[p1] which is in part1 (N[p1] < N[0]) and N[p3] in part3 (N[0] < N[p3]).

if no number in the 3 part is more then 51, return false.

otherwise, if part2 >= 51, ok N[0] is the one.
if either part1 or part3 is more than 51, here suppose part3 is more than 51. since N[p3] in part3( N[0] < N[p3]), we can use both of values N[0] and N[p3] to go through the array again. this time, just focus on how many are between N[0] and N[p3], how many are equal to N[p3] and how many are more then N[p3].

Now we have new three parts (par1, between N[0] and N[p3], part2 equal to N[p3], part3, more than N[p3]).
at the same time, don't forget to pick up a value in part1 and part3 as new N[p1] and new N[p3] for next time compare.
continue above process, until find or fail.

[小童012]

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[ximengchang]

学习

[steal8275756]

[Quote=引用 106 楼 suiyili 的回复:]
Now we have new three parts (par1, between N[0] and N[p3], part2 equal to N[p3], part3, more than N[p3]).
at the same time, don't forget to pick up a value in part1 and part3 as new N[p1] and new N[p3] for next time compare.
continue above process, until find or fail.[/Quote]
红字部分应该是less than吧

[suiyili]

yes, steal, you are absolutely right, thx.

[remagon]

[Quote=引用 106 楼 suiyili 的回复:]
there is some mistake above, here is the correct one.

If an array N has S elements, then 51% means that there must be S*0.51 number of same elements.

Here, to be simple, suppose we have 100 elements, then ramdomly find an element, ok, we use the first element N[0], then, compared with N[0], go through every item in the array, after that we can get how many elements which are more than N[…
[/Quote]
递归了，而且要使用临时数组

[qq441733920]

学习.

[zexin1000]

路过，觉得很有挑战性。

[suiyili]

to remagon:

there is no recursive in my method, the two values N[p1] and N[p3] should be able to updated in a loop.

[Mydal]

12楼的算法很不错啊 挺符合题目的

[fengquansheng]

楼主：给你写个好算法你也领悟不了。

[--小F--]

学习一下

[alleneros1984]

3:Code for Maximum Subset Sum

#include "limits.h"
#include "stdio.h"

int maxsub(int* arr, int len){

int i, maxsum = INT_MIN, sum=0;
int beg = 0, end = 0, pos = 0;

for (i=0; i<len; i++){

sum += arr[i];
if(sum > maxsum){

maxsum = sum;
beg = pos;
end = i;
}
if(sum < 0){

sum = 0;
pos = i+1;
}
}

for(i=beg; i<=end; i++)
printf("%d \t",arr[i]);
printf("\n");
return maxsum;
}

[xnhsdp]

学习。。。。。。。。。。。。。