【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