First Position of Target

First position of Target.png

===================== 解題思路 =====================

BS 方式尋找到 target, 但是這次可能會有重複的數字存在 所以需要找到的是index 最小的target 位置 所以在找到 target 之後並不中斷 loop, 繼續壓縮範圍往 target 集中 最後在 while loop 設定的條件下結束時 照順序檢查 left 跟 right 是否為 target (left 優先) 如果都沒有回傳 -1

===================== C++ code ====================

<pre><code>
class Solution {

public:

/**
 * @param nums: The integer array.
 * @param target: Target number to find.
 * @return: The first position of target. Position starts from 0. 
 */
int binarySearch(vector<int> &array, int target) {
    // write your code here
    int left = 0, right = array.size() - 1;
    while(left + 1 < right)
    {
        int mid = left + (right - left) / 2;
        if(array[mid] >= target) right = mid;
        else left = mid;
    }
    if(array[left] == target) return left;
    if(array[right] == target) return right;
    return -1;
}

};
<code><pre>