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  • 【leetcode】885. Boats to Save People

    题目如下:

    解题思路:本题可以采用贪心算法,因为每条船最多只能坐两人,所以在选定其中一人的情况下,再选择第二个人使得两人的体重最接近limit。考虑到人的总数最大是50000,而每个人的体重最大是30000,因此会有很多人体重一样。这样可以用一个集合set保存体重,再用字典保存每个体重对应的人的数量,可以减少运算量。最后对set进行升序排序,再用双指针的方法,用low指针指向set头部,high指针指向set尾部,分别比较set[low]和set[high]的值,有如下情况:

    a. set[low] + set[high] > limit: 表示set[high]的体重人只能一人坐一船,得出boats += dic[set[high]], high -= 1;

    b. set[low] + set[high] <= high: 表示可以共坐一船,但这里又要考虑  dic[set[high]] 与  dic[set[low]] 的大小:

       b1. dic[set[high]]  > dic[set[low]]  : 得出 boats += dic[set[low]], low += 1;

       b2.dic[set[high]]  < dic[set[low]]  : 得出 boats += dic[set[high]], high -= 1;

       b3:dic[set[high]]  == dic[set[low]]  : 得出 boats += dic[set[high]], high -= 1,low += 1;

    最后当low == high的时候,表示只剩下相同体重的人,这时判断这些人能否共坐一船,得出最终boats的总数。

    代码如下:

    class Solution(object):
        def numRescueBoats(self, people, limit):
            """
            :type people: List[int]
            :type limit: int
            :rtype: int
            """
            dic = {}
            puniq = []
            for i in people:
                if i not in dic:
                    dic[i] = 1
                    puniq.append(i)
                else:
                    dic[i] += 1
            puniq.sort()
            boats = 0
            low = 0
            high = len(puniq) - 1
            while low < high:
                if puniq[low] + puniq[high] > limit:
                    boats += dic[puniq[high]]
                    dic[puniq[high]] = 0
                    high -= 1
                else:
                    if dic[puniq[low]] > dic[puniq[high]]:
                        boats += dic[puniq[high]]
                        dic[puniq[low]] -= dic[puniq[high]]
                        dic[puniq[high]] = 0
                        high -= 1
                    elif dic[puniq[low]] < dic[puniq[high]]:
                        boats += dic[puniq[low]]
                        dic[puniq[high]] -= dic[puniq[low]]
                        dic[puniq[low]] = 0
                        low += 1
                    else:
                        boats += dic[puniq[high]]
                        dic[puniq[high]] = 0
                        dic[puniq[low]] = 0
                        low += 1
                        high -= 1
            if limit >= puniq[high]*2:
                boats += ((dic[puniq[high]]/2) + (dic[puniq[high]]%2))
            else:
                boats += dic[puniq[high]]
            #boats += dic[puniq[low]]
    
            return boats
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  • 原文地址:https://www.cnblogs.com/seyjs/p/9428419.html
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