# Use Leetcode to familiarize yourself with dynamic planning

Source: Internet
Author: User
Introduction

General thoughts on solving the problem of dynamic regulation

decompose the original problem into sub-problems

The original problem is decomposed into several sub-problems, the sub-problem and the original problem form the same or similar, but the size becomes smaller. Sub-problems are solved, the original problem is solved (digital triangle example).
Once the solution of a sub-problem is found, it is saved, so each sub-problem is solved only once.

Determine status

When solving problems with dynamic programming, we often refer to a set of variables related to sub-problems as a "state". A "state" corresponds to one or more sub-problems, the so-called "state" under the "value" is the "state" corresponding to the sub-problem of the solution.
A collection of all "states" that constitute the "state space" of the problem. The size of the state space is directly related to the time complexity of solving the problem with dynamic planning.

determine the value of some initial state (boundary State)

determining the state transition equation Leetcode Simple Topic Introduction Dynamic Planning

```public class Dp_algorithm {//================= Dynamic planning first question =====================================///** * ======== * ======================== Topic ========================= * 198. House Robber * is a professional robber planning to rob houses along a street. Each house have a certain amount of money stashed, the only constraint stopping all from robbing each of the them are that Adjac ENT houses have security system connected and it would automatically contact the police if the adjacent houses were broken
Into on the same night.  * *given a list of non-negative integers representing the amount of money in each house, determine the maximum amount
The money you can rob tonight without alerting the police. * * * * */** * Solution One: * Brute Force search * We assume the first step from the last. * There are two ways to do this: * n-1 is the second step only (N-3,n-4,....
* N-2 The second step only (N-4,,n-5,...)
* */public static int solve1 (int idx,int[] nums) {if (idx<0) {return 0;
}        Return Math.max (Nums[idx]+solve1 (idx-2,nums), solve1 (idx-1,nums)); }/** * Solution II: * Algorithm after optimization * We assume that the first step from the last one is two ways of thinking: * N-1 is the second step only (N-3,n-4,....
* N-2 The second step only (N-4,,n-5,...) * By observing we can see n-4,n-5 repeated calculations, then we will save the calculated results, * So save time * *///public static int[] public static int Sol
Ve2 (int idx,int[] nums) {if (idx<0) {return 0;
} if (results[idx]>0) {return RESULTS[IDX];
} Results[idx] = Math.max (Nums[idx]+solve2 (idx-2,nums), Solve2 (idx-1,nums));
return RESULTS[IDX];
} static int[] results;
public static void Main (string[] args) {int[] nums = new Int[3];
Nums[0] = 1;
NUMS[1] = 2;
NUMS[2] = 3;
results = new Int[nums.length];
for (int i = 0; i < nums.length; i++) {Results[i] =-1;
} int result = Solve2 (nums.length-1, nums);   System.out.print (result); }
}```
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