第一章 习题集 – 图领1000题

#include <stdio.h>

int count_ones(int nums[], int n) {
    int left = 0;
    int right = n - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == 1) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return right + 1;
}

int main() {
    // 示例1: 正常情况
    int arr1[] = {1, 1, 1, 0, 0};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    printf("数组1中1的个数为: %d\n", count_ones(arr1, n1));  // 应输出3

    // 示例2: 全1数组
    int arr2[] = {1, 1, 1, 1};
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    printf("数组2中1的个数为: %d\n", count_ones(arr2, n2));  // 应输出4

    // 示例3: 全0数组
    int arr3[] = {0, 0, 0};
    int n3 = sizeof(arr3) / sizeof(arr3[0]);
    printf("数组3中1的个数为: %d\n", count_ones(arr3, n3));  // 应输出0

    // 示例4: 单个1
    int arr4[] = {1};
    int n4 = sizeof(arr4) / sizeof(arr4[0]);
    printf("数组4中1的个数为: %d\n", count_ones(arr4, n4));  // 应输出1

    // 示例5: 单个0
    int arr5[] = {0};
    int n5 = sizeof(arr5) / sizeof(arr5[0]);
    printf("数组5中1的个数为: %d\n", count_ones(arr5, n5));  // 应输出0

    return 0;
}