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  • 毕业生面试题

    十大经典排序算法
    https://www.runoob.com/w3cnote/ten-sorting-algorithm.html
    Java最新常见面试题 + 答案汇总
    https://blog.csdn.net/fangchao2011/article/details/89203535

    11、输入一个递增排序的数组nums和一个数字target,在数组中查找两个数,使得它们的和正好是target。如果有多对数字的和等于target,输出全部组合。30

    要求: 要考虑时间复杂度和空间复杂度
    如果数组中不存在目标值,返回 [-1, -1]。
    示例 1:
    输入: nums = [1,2,4,7,8,11,15], target = 15
    输出: [4,11], [7,8]
    示例 2:
    输入: nums = [5,7,9,10,13], target = 23
    输出: [-1,-1]

    12、输入一个正数target,打印出所有和为target的连续正数序列(至少含有两个数)。例如输入15,
    由于1+2+3+4+5=4+5+6=7+8=15,所以结果打印出3个连续序列1~5、4~6和7~8。35

    public static void main(String[] args) {

    //num=[1,2,4,7,8,11,15]  target=15  输出  [4,11]  [7,8]              ,//11

    //findContinueSequence(15);

    int[] num= new int[] {1,3,5,6,8,9};   //输入一个正数 target =13  ,//12
    System.out.println(FindNumbersWithSum(num,13));
    }

    public static ArrayList<Integer> FindNumbersWithSum(int[]num, Integer target) {
    int p = 0;
    int q = num.length- 1;
    ArrayList<Integer> rst = new ArrayList<Integer>();

    while (p < q) {
    if (num[p] + num[q] == target) {
    rst.add(num[p]);
    rst.add(num[q]);
    break;
    } else if (num[p] + num[q] > target) {
    q--;
    } else {
    p++;
    }
    }
    return rst;
    }


    private static void findContinueSequence(int target) {
    if (target < 3) {
    return;
    }
    int start = 1;
    int end = 2;
    int mid = (1 + target) / 2;
    int sum = start + end;
    while (start < mid) {
    if (sum == target) {
    printNum(start, end);
    }
    while (sum > target && start < mid) {
    sum -= start;
    start++;
    if (sum == target) {
    printNum(start, end);
    }
    }
    end++;
    sum += end;

    }
    }

    private static void printNum(int start, int end) {
    for (int i = start; i <= end; i++) {
    System.out.print(i + " ");
    }
    System.out.println();

    }

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  • 原文地址:https://www.cnblogs.com/kelelipeng/p/12331820.html
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