29. 数组中出现次数超过一半的数字.

方法a. 排序取中 O(nlogn).

方法b. partition 函数分割找中位数 >=O(n).

方法c. 设计数变量，扫描一遍。 O(n).

#include <stdio.h>
int getNumber(int data[], int length){
/*  if(checkInvalidArray(data, length)) return 0;  */
    int count = 1, value = data[0];
    for(int i = 1; i < length; ++i)
    {
        if(count == 0){
            value = data[i];
        }else if(data[i] == value){
            ++count;
        }else
            --count;
    }
    return value;
}
int main(){
    int numbers[] = {2, 2, 2, 2, 6, 6, 6, 6, 6};
    int value = getNumber(numbers, sizeof(numbers) / 4);
/*  if(value != 0 || !checkInvalidArray(data, length))  */
    printf("%d\n", value);
    return 0;
}

30. 最小的 k 个数

a. partition 函数找到第 k 个数. >=O(n)

#include <stdio.h>
int partition(int data[], int low, int high){
    int value = data[low];
    while(low < high){
        while(low < high && data[high] >= value) --high;
        data[low] = data[high];
        while(low < high && data[low] <= value) ++low;
        data[high] = data[low];
    }
    data[low] = value;
    return low;
}
void getKNumber(int input[], int length, int out[], int k){
    if(!input || !out || length < 1 || k > length || k < 1) return;
    int low = 0, high = length - 1, index;
    do{
        index = partition(input, low, high);
        if(index < k-1) low = index + 1;
        else if(index > k-1) high = index - 1;
    }while(index != k-1);
    for(int i = 0; i < k; ++i)
        out[i] = input[i];
}
int main(){
    int numbers[10] = {3, 5, 2, 6, 7, 4, 9, 1, 2, 6};
    int k = 5;
    getKNumber(numbers, 10, numbers, k);
    for(int i = 0; i < k; ++i)
        printf("%-3d", numbers[i]);
    printf("\n");
    return 0;
}

b. 构造k 个元素的大顶堆

#include <stdio.h>
void HeapAdjust(int data[], int endIndex, int father){
    if(!data || endIndex < 0 || father > endIndex || father < 0) return;
    int value = data[father]; // set data[0] to save the value of original father
    for(int child = 2*father+1; child <= endIndex; child = 2*father+1){
        if(child < endIndex && data[child] < data[child+1]) ++child;
        if(data[child] < value) break;
        else data[father] = data[child];
        father = child;
    }
    data[father] = value;
}
void getKNumber(int input[], int length, int out[], int k){
    if(!input || !out || length < 1 || k > length || k < 1) return;
    for(int i = 0; i < k; ++i)
        out[i] = input[i];
    for(int i = k/2-1; i >= 0; --i)
        HeapAdjust(out, k-1, i);
    for(int i = k; i < length; ++i){
        if(input[i] < out[0]){
            out[0] = input[i];
            HeapAdjust(out, k-1, 0);
        }
    }
}
int main(){
    int numbers[10] = {3, 5, 2, 6, 7, 4, 9, 1, 2, 6};
    enum{ k = 1};
    int out[k+1] = {0};
    getKNumber(numbers, 10, out, k);
    for(int i = k-1; i >= 0; --i){
        int tem = out[i];
        out[i] = out[0];
        out[0] = tem;
        printf("%-3d", out[i]);
        HeapAdjust(out, i-1, 0); // DESC
    }
    printf("\n");
    return 0;
}

31. 连续子数组的最大和

#include <stdio.h>
bool Invalid_Input = false;
int getKNumber(int data[], int length){
    Invalid_Input = false;
    if(data == NULL || length < 1) {
        Invalid_Input = true;
        return 0;
    }
    int maxSum = 0x80000000;
    int curSum = 0;
    for(int i = 0; i < length; ++i){
        if(curSum < 0) curSum = data[i];
        else curSum += data[i];
        if(curSum > maxSum) maxSum = curSum;
    }
    return maxSum;
}
int main(){
    int numbers[] = {1,-2, 3, 10, -4, 7, 2, -5, -2, 4, -5, 4};
    int maxSum = getKNumber(numbers, sizeof(numbers)/4);
    if(!Invalid_Input)
        printf("%d\n", maxSum);
    return 0;
}

优质内容筛选与推荐>>
1、李嘉诚用人之道的关键
2、（线段树） 单点更新，区间查询最值
3、判断一个字符串（超过80个字符）是否是回文结构（正序和逆序相同）
4、PAT_A1125#Chain the Ropes
5、JQuery 获取json数据$.getJSON方法的实例代码

赞赏

长按二维码向我转账

受苹果公司新规定影响，微信 iOS 版的赞赏功能被关闭，可通过二维码转账支持公众号。

阅读

已发送

发送中