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  • leetcode 605. Can Place Flowers

    Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots - they would compete for water and both would die.

    Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule.

    Example 1:

    Input: flowerbed = [1,0,0,0,1], n = 1
    Output: True
    

    Example 2:

    Input: flowerbed = [1,0,0,0,1], n = 2
    Output: False
    

    Note:

    1. The input array won't violate no-adjacent-flowers rule.
    2. The input array size is in the range of [1, 20000].
    3. n is a non-negative integer which won't exceed the input array size.

    我的方法是枚举

    class Solution {
    public:
        bool canPlaceFlowers(vector<int>& flowerbed, int n) {
            int len = flowerbed.size();
            int num = 0;
            for (int i = 0; i < len; i +=2) {
                if (flowerbed[i] == 0) {
                    if ( (i+1 >=len ||flowerbed[i + 1] == 0) && (i-1 < 0 ||flowerbed[i - 1] == 0) ) num++;
                }
            }
            if (num >= n) return true;
            num = 0;
            for (int i = 1; i < len; i += 2) {
                if (flowerbed[i] == 0) {
                    if ((i+1 >= len || flowerbed[i + 1] ==0) && (i-1 < 0 || flowerbed[i - 1] == 0)) num++;
                }
            }
            if (num >= n) return true;
            return false;
        }
    };

    网上看了个不错的思路如下:

    class Solution {
    public:
        bool canPlaceFlowers(vector<int>& flowerbed, int n) {
            int result = 0;
            int count = 1;
            // edge case: left and right
            for (int i = 0; i < flowerbed.size(); i++) {
                if (flowerbed[i] == 0) {
                    count++;
                } else {
                    result += (count - 1) / 2;
                    count = 0;
                }
            }
            if (count != 0) {
                result += count / 2;
            }
            return result >= n;
        }
    };
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  • 原文地址:https://www.cnblogs.com/pk28/p/7253881.html
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