LintCode-First Position of Target

First Position of Target

Description

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

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

Challenge

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

• 题目模板

  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) {
        // write your code here
    }
  };
  

• 题目大意

  给你个数组，再给你一个数，找出这个数在这个数组中的最低位置。要求时间复杂度为O(logn)。

• 大概思路

  题目已经给的提示很明显了，二分查找，或者高级一点三分查找也可以。

  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) {
        // write your code here
        int le = 0, ri = nums.size()-1;
        int min = -1;
        int first = 1;
        while(le <= ri){
          int mid = (le + ri) / 2;
          if(nums[mid] == target){
            if(first){
              first = 0;
              min = mid;
            }
            else
              min = min < mid ? min : mid;
            ri = mid - 1;
          }
          else if(nums[mid] > target)
            ri = mid - 1;
          else
            le = mid + 1;
        }
        return min;
    }
  };

• 细节方面

  题目中要求是最低位置，所以我们还要维护一个最低位置的变量，而不是像以前一样找到就行，那么改变就是，当nums[mid] == target时不直接返回，而是更新维护的最小位置值，然后ri = mid-1就可以了，这里因为是维护最小值，所以再向左面找，动ri而不是le。

• 题目链接：https://www.lintcode.com/problem/first-position-of-target/description