四道简单DP

　　DP类题目找到子问题（状态），然后找到转移方程，就OK

#dp
#likes matrixchain
#according to two point's distance to recurrence
class Solution:
    # @return a string
    def longestPalindrome(self, s):
        length = len(s)
        p = [[0 for col in range(length)] for row in range(length)]
        for i in range(len(s)):
            p[i][i] = 1
            if(i<(length-1) and s[i] == s[i+1]):
                maxlenth = 2
                p[i][i+1] = 1
                start = i
        for i in range(3,length+1):
            for j in range(length-i+1):
                k = j + i - 1
                if(s[j]==s[k] and p[j+1][k-1] == 1):
                    maxlenth = i
                    p[j][k] = 1
                    start = j
        return s[start:start+maxlenth]


if __name__ == '__main__':
    s = Solution()
    test_str = 'abcba'
    print s.longestPalindrome(test_str)

#dp least number coins
#one:overlapping subproblem
#two:optimal substructure
class Solution:
    # @return a string
    def least_number_coin(self, s, v):
        #first initialization
        min_number = [1000000]*(s+1)
        min_number[0] = 0
        #And then,recurrence
        #time need O(n^2),and additional time O(n) to save the subproblem's temp solution
        for i in range(1,s+1):
            for j in range(len(v)):
                print i,v[j]
                if(v[j]<=i and (min_number[i-v[j]]+1 < min_number[i])):
                    min_number[i] = min_number[i-v[j]]+1
        print min_number
        return min_number[s]
if __name__ == '__main__':
    s = Solution()
    money = 11
    v = [1,3,5]
    print s.least_number_coin(money,v)

#dp
#time O(n^2),addtional O(n) to save temp result
class Solution:
    # @return a string
    def LIS(self,v):
        print v
        d = [1]*(len(v))
        print d
        for i in range(len(v)):
            for j in range(i):
                print i,j
                if (v[j]<=v[i] and d[j]+1>d[i]):
                    d[i] = d[j]+1
        print d
        print max(d)
if __name__ == '__main__':
    s = Solution()
    v = [5,3,4,8,6,7]
    s.LIS(v)

#dp
class Solution:
    # @return a string
    def max_number_apples(self, apples, n, m):
        print apples
        s = [[0 for col in range(m)] for row in range(n)]
        for i in range(n):
            for j in range(m):
                if(i==0 and j==0):
                    s[i][j] = apples[0][0]
                elif(i==0 and j>0):
                    s[i][j] = s[i][j-1]+apples[i][j]
                elif(j==0 and i>0):
                    s[i][j] = s[i-1][j] + apples[i][j]
                else:
                    if(s[i-1][j]>s[i][j-1]):
                        s[i][j] = s[i-1][j] + apples[i][j]
                    else:
                        s[i][j] = s[i][j-1] + apples[i][j]
        print s
        return s[n-1][m-1]
if __name__ == '__main__':
    s = Solution()
    n = 3
    m = 4
    apples = [[0 for col in range(m)] for row in range(n)]
    k = 1
    for i in range(n):
        for j in range(m):
            apples[i][j] = k
        k = k + 1
    s.max_number_apples(apples,n,m)

　　one:initialization（初始化）

　　two:recurrence（递推）

　　多项式时间，需要额外的存储空间