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  • [array] leetcode

    leetcode - 34. Search for a Range - Medium

    descrition

    Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

    Your algorithm's runtime complexity must be in the order of O(log n).

    If the target is not found in the array, return [-1, -1].

    For example,
    Given [5, 7, 7, 8, 8, 10] and target value 8,
    return [3, 4].

    解析

    对于有序数组的查找问题,基本上都可以使用折半查找的思路。再者题目的要求复杂度是 O(log n),因此我们需要两次折半查找。如果我们只用一次折半查找,找到数组 target 出现的任意一个位置,然后线性遍历找到最左边和最右,需要的复杂度是 O(n)。

    具体实现代码如下。

    code

    
    #include <iostream>
    #include <vector>
    #include <algorithm>
    
    using namespace std;
    
    class Solution{
    public:
    	vector<int> searchRange(vector<int>& nums, int target) {
    		vector<int> ans(2, -1);
    		int ileft = binarySearchLeftMost(nums, target);
    		if(ileft < 0)
    			return ans;
    		
    		int iright = binarySearchRightMost(nums, target);
    
    		ans[0] = ileft;
    		ans[1] = iright;
    
    		return ans;
    	}
    
    	int binarySearchLeftMost(vector<int>& nums, int target){
    		int ans = -1; // save the left most index in nums which equal to target
    		int ileft = 0, iright = nums.size() - 1;
    
    		while(ileft <= iright){
    			int imid = ileft + (iright - ileft) / 2;
    			if(nums[imid] == target){
    				ans = imid;
    				iright = imid - 1;
    			}else if (nums[imid] < target){
    				ileft = imid + 1;
    			}else{
    				// nums[imid] > target
    				iright = imid - 1;
    			}
    		}
    
    		return ans;
    	}
    
    	int binarySearchRightMost(vector<int>& nums, int target){
    		int ans = -1;
    		int ileft = 0, iright = nums.size() - 1;
    
    		while(ileft <= iright){
    			int imid = ileft + (iright - ileft) / 2;
    			if(nums[imid] == target){
    				ans = imid;
    				ileft = imid + 1;
    			}else if (nums[imid] < target){
    				ileft = imid + 1;
    			}else {
    				// nums[imid] > target
    				iright = imid - 1;
    			}
    		}
    
    		return ans;
    	}
    
    
    };
    
    int main()
    {
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/fanling999/p/7841600.html
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