关于java:聚焦市场开发热门技术-手把手带你开发商业级社交App无密

38次阅读

共计 1510 个字符,预计需要花费 4 分钟才能阅读完成。

download: 聚焦市场开发热门技术 手把手带你开发商业级社交 App

实例

include<stdio.h>

define MAX 20

//#define SHOWPASS

define BASE 10

void print(int *a, int n) {
int i;
for (i = 0; i < n; i++) {

printf("%d\t", a[i]);

}
}

void radixsort(int *a, int n) {
int i, b[MAX], m = a[0], exp = 1;

for (i = 1; i < n; i++) {

if (a[i] > m) {m = a[i];
}

}

while (m / exp > 0) {

int bucket[BASE] = {0};

for (i = 0; i < n; i++) {bucket[(a[i] / exp) % BASE]++;
}

for (i = 1; i < BASE; i++) {bucket[i] += bucket[i - 1];
}

for (i = n - 1; i >= 0; i--) {b[--bucket[(a[i] / exp) % BASE]] = a[i];
}

for (i = 0; i < n; i++) {a[i] = b[i];
}

exp *= BASE;

ifdef SHOWPASS

printf("\nPASS   :");
print(a, n);

endif

}
}

int main() {
int arr[MAX];
int i, n;

printf(“Enter total elements (n <= %d) : “, MAX);
scanf(“%d”, &n);
n = n < MAX ? n : MAX;

printf(“Enter %d Elements : “, n);
for (i = 0; i < n; i++) {

scanf("%d", &arr[i]);

}

printf(“\nARRAY : “);
print(&arr[0], n);

radixsort(&arr[0], n);

printf(“\nSORTED : “);
print(&arr[0], n);
printf(“\n”);

return 0;
}
Lua
实例
— 获取表中位数
local maxBit = function (tt)

local weight = 10;      -- 十進制
local bit = 1;

for k, v in pairs(tt) do
    while v >= weight do
        weight = weight * 10;
        bit = bit + 1;  
    end
end
return bit;

end
— 基数排序
local radixSort = function (tt)

local maxbit = maxBit(tt);

local bucket = {};
local temp = {};
local radix = 1;
for i = 1, maxbit do
    for j = 1, 10 do
        bucket[j] = 0;      --- 清空桶
    end
    for k, v in pairs(tt) do
        local remainder = math.floor((v / radix)) % 10 + 1;    
        bucket[remainder] = bucket[remainder] + 1;      -- 每個桶數量自動减少 1
    end
   
    for j = 2, 10 do
        bucket[j] = bucket[j - 1] + bucket[j];  -- 每个桶的数量 = 以前桶数量和 + 自个数量
    end
    -- 按照桶的地位,排序 -- 这个是桶式排序,必须使用倒序,因为排序方法是从小到大,次序下来,会出现大的在小的下面清空。for k = #tt, 1, -1 do
        local remainder = math.floor((tt[k] / radix)) % 10 + 1;
        temp[bucket[remainder]] = tt[k];
        bucket[remainder] = bucket[remainder] - 1;
    end
    for k, v in pairs(temp) do
        tt[k] = v;
    end
    radix = radix * 10;
end

end;

正文完
 0