Typical problem of dynamic programming 03: Maximum number-to-Number Difference (or minimum number-to-Number Difference) in the array)

Question: In the array, subtract the number on the right of the number to get a number pair. Calculate the maximum value of the difference between all number pairs. For example, in the array {2, 4, 1, 16, 7, 5, 11, 9}, the maximum value of the number pair is 11, which is the result of 16 minus 5.

Reference: http://zhedahht.blog.163.com/blog/static/2541117420116135376632/

Forum: http://stackoverflow.com/questions/11858790/find-the-largest-possible-difference-in-an-array-with-the-smaller-integer-occuri

[Analysis] top-down analysis, where diff [I] records 0 ~ Max [I] records 0 ~ The maximum number of I, the State recursive equation is:

Diff [I + 1] = math. Max (MAX [I]-array [I + 1], diff [I])

For bottom-up analysis, the first test is Max [1] = 2, diff [1] = 2-4 =-2. Max [2] = 4, diff [2] = max (4-1, diff [1]) = 3. Max [3] = 4, diff [3] = max (4-16, diff [2]) = 3 .......

The algorithm complexity is O (n ).

/*** Creation Time: 5:27:46, January 1, September 5, 2014 * Project name: test * @ author Cao yanfeng * @ since JDK 1.6.0 _ 21 * class description: dynamic Programming resolves the maximum number of pairs in the array */public class maxdiffofarray {/*** @ paramargs */public static void main (string [] ARGs) {// todo auto-generated method stub int [] array = {, 0,-1}; system. out. println (getmaxdiff (array);} public static int getmaxdiff (INT [] array) {int length = array. length; If (length = 1) {return-999;} If (length = 2) {return array [2]-array [1];} int max = array [0]; int maxdiff = max-array [1]; for (INT I = 2; I <length; I ++) {If (array [1]> MAX) {max = array [I-1];} int currentdif = max-array [I]; If (currentdif> maxdiff) {maxdiff = currentdif ;}} return maxdiff ;}}

Typical problem of dynamic programming 03: Maximum number-to-Number Difference (or minimum number-to-Number Difference) in the array)