Binary Search

• 專題1.二分查詢
• 有序中查詢兩個元素

001.Binary Search

// template 1
// end the loop ,the  left==right
int binarySearch(vector<int>& nums, int target){
    if(nums.size() == 0)
       return -1;

    int left = 0, right = nums.size();
    while(left < right){
    // Prevent (left + right) overflow
       int mid = left + (right - left) / 2;
       if(nums[mid] == target)
            { return mid; }
       else if(nums[mid] < target)
            { left = mid + 1; }
       else 
            { right = mid; }
    }

  // Post-processing:
  // End Condition: left == right
    if(left != nums.size() && nums[left] == target)
        return left;
    return -1;
}


// template 2
// left+1==right
// 這種方法應用到題目中就是不知道該怎麼停止，確定就直接找到兩個重複元素
// 就是直接指標
int binarySearch(vector<int>& nums, int target){
    if (nums.size() == 0)
        return -1;

    int left = 0, right = nums.size() - 1;
    while (left + 1 < right){
        // Prevent (left + right) overflow
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid;
        } else {
            right = mid;
        }
    }

    // Post-processing:
    // End Condition: left + 1 == right
    if(nums[left] == target) return left;
    if(nums[right] == target) return right;
    return -1;
}
 

search for a range

// 我感覺題目有問題
// 而且，這個程式碼還是不是理解的很透徹
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res(2,-1);
        if(nums.size()<=0)
            return res;
        int start=0;
        int end=nums.size()-1;
        while(start+1<end){
            int mid=start+(end-start)/2;
                
            if(nums[mid]<target)
                start=mid+1;
            else
                end=mid;
        }
        if(nums[start]==target && nums[end]==target)
        {
            res[0]=start;
            res[1]=end;
        }
           
        return res;
        
    }
};
 

三種模版之間的比較

Note: The templates and their differences have been colored coded below.

Some problems can be implemented using multiple templates, but as you practice more, you will notice that some templates are more suited for certain problems than others.

These 3 templates differ by their:

• left, mid, right index assignments
• loop or recursive termination condition
• necessity of post-processing