Dynamic Programming之Longest Increasing Subsequence (LIS)问题
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Longest Increasing Subsequence(LIS)问题属于经典的DP问题。
其DP是模型:
对于序列a[total],使用lis[i]表示前i位包含第i个元素的LIS,因此有转移方程
if(a[j] > a[i])
lis[j] = max( list[i] + 1, list[j])
算法描述如下:
for i = 1 to total-1
for j = i+1 to total
if a[j] > a[i] then
if lis[i] + 1 > lis[j] then
lis[j] = lis[i] + 1
for j = i+1 to total
if a[j] > a[i] then
if lis[i] + 1 > lis[j] then
lis[j] = lis[i] + 1
这个算法的时间复杂度为O(n2)。举个例子来演示这个算法,取一个序列初始化如下:
a = {9, 5, 2, 8, 7, 3, 1, 6, 4}
lis = {1, 1, 1, 1, 1, 1, 1, 1, 1}//初始化时lis全部要初始化为1,这与以往经典的DP问题不同
lis = {1, 1, 1, 1, 1, 1, 1, 1, 1}//初始化时lis全部要初始化为1,这与以往经典的DP问题不同
然后使用这个算法,可以得到如下的结果:
a = {9, 5, 2, 8, 7, 3, 1, 6, 4}
lis = {1, 1, 1, 2, 2, 2, 1, 3, 3}
lis = {1, 1, 1, 2, 2, 2, 1, 3, 3}
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