Coursera Algorithms, Part 1 Week 1: Analysis of Algorithm
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这一部分主要是算法理论上的分析,包括时间和空间的复杂度,仅仅找了一个比较有意思的例子来设计算法。
问题:给定N个不同的整数,计算有多少这样的组合使得其中三个数之和为0。
暴力的O(N^3)算法就不说了,下面来看个另外一种的算法。
(1)对于这N个不同的数,先将这N个数进行排序,O(N*lgN)
(2)找出所有的两个数的组合(ai, aj) 0<=i, j<=N-1且i!=j,对每一对(ai, aj),计算-(ai+aj)的值,O(N^2)
(3)对于每一个-(ai+aj)的值,在已经排序的序列中二分查找,确定是否存在该值,存在的话,计数加一,O(lgN)
(4)为避免重复计数,可以增加判断条件,O(1)
算法的复杂度从O(N^3)降到O(N^2*lgN)
还是那句话,每天积累一点点,总会有牛逼的一天的!
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