线性查找法时间复杂度为0(n)
    Java实现: 

    1. /**
    2. * @param  nums   数组 
    3. * @param  target 目标参数 
    4. * @return  目标参数下标 
    5. * 
    6. * 从下标0开始循环,若当前下标的值等于目标参数,返回当前下标,若循环结束,为找到目标参数,则返回-1
    7. */
    8. public static <E> Integer search(E[] nums, E target) {
    9.         int length = nums.length;
    10.         for (int i = 0; i < length; i++) {
    11.             if (nums[i].equals(target)) {
    12.                 return i;
    13.             }
    14.         }
    15.         return -1;
    16.     }
    17. public static void main(String[] args) {
    18.     Integer[] nums = {5, 2, 4, 6, 1, 3};
    19.     System.out.println(search(nums, 10));
    20.     System.out.println(search(nums, 6));
    21. }

    go实现:

    1. func main() {
    2.    var nums = []int{5, 2, 4, 6, 1, 3}
    3.    fmt.Println(linearSearch(nums, 6))
    4.    fmt.Println(linearSearch(nums, 7))
    5. }
    6. func linearSearch(nums []int, b int) int {
    7.    length := len(nums)
    8.    for i := 0; i < length; i++ {
    9.       if nums[i] == b {
    10.          return i
    11.       }
    12.    }
    13.    return -1
    14. }

    项目demo