LintCode 14: First Position of Target

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于 2019-05-26 01:33:12 发布

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分类专栏： algorithm-design 文章标签： LintCode

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博客围绕在给定的升序排序数组中查找目标数的首个索引展开，要求时间复杂度为O(log n)，若目标数不存在则返回 -1。给出了多个示例，还提出若数组元素数量大于2^32代码能否正常工作的挑战，并给出了类似二分查找的解法。

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First Position of Target
中文English
For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example
Example 1:
Input: [1,4,4,5,7,7,8,9,9,10]，1
Output: 0

Explanation: 
the first index of  1 is 0.

Example 2:
Input: [1, 2, 3, 3, 4, 5, 10]，3
Output: 2

Explanation: 
the first index of 3 is 2.

Example 3:
Input: [1, 2, 3, 3, 4, 5, 10]，6
Output: -1

Explanation: 
Not exist 6 in array.

Challenge
If the count of numbers is bigger than 2^32, can your code work properly?

解法1：类似Binary Search。
代码如下：

class Solution {
public:
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    int binarySearch(vector<int> &nums, int target) {
        int n = nums.size();
        if (n == 0) return -1;
        
        int start = 0, end = n - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] < target) {
                start = mid;
            } else if (nums[mid] >= target) {
                end = mid;
            }
        }
        
        if (nums[start] == target) return start;
        if (nums[end] == target) return end;
        return -1;
    }
};