LeetCode-堆-Easy

记录了初步解题思路 以及本地实现代码；并不一定为最优 也希望大家能一起探讨 一起进步

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703.kth-largest-element-in-a-stream 数据流中的第K大元素

解题思路：在heap中保留k个数 来一个新的数时与heap中的第一个数 及第K大元素的数比较

import heapq

class KthLargest(object):

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.nums = nums
        self.size = len(self.nums)
        self.k = k
        heapq.heapify(self.nums)
        while self.size > k:
            heapq.heappop(self.nums)
            self.size -= 1
            
    def add(self,val):
        """
        :type val: int
        :rtype: int
        """
        if self.size<self.k:
            heapq.heappush(self.nums,val)
            self.size+=1
        elif val>self.nums[0]:
            heapq.heapreplace(self.nums,val)
        return self.nums[0]

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743.network-delay-time 网络延迟时间

解题思路：

def networkDelayTime(times, N, K):
    """
    :type times: List[List[int]]
    :type N: int
    :type K: int
    :rtype: int
    """
    INF = 0x7FFFFFFF
    gradic = {}
    resdic = {}
    for i in  range(len(times)):
        u,v,w = times[i]
        ul = gradic.get(u,[])
        ul.append((v,w))
        gradic[u] = ul[:]
        
    line = [K]
    resdic[K]=0
    
    while line:
        k = line.pop()
        tmppath = resdic[k]
        tmpl = gradic.get(k,[])
        for tv,tw in tmpl:
            imapath = resdic.get(tv,INF)
            if tw+tmppath < imapath:
                resdic[tv]=tw+tmppath
                line.append(tv)
    if len(resdic)<N:
        return -1
    return max(resdic.values())

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