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MinAdjustmentCost_Class.java
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243 lines (200 loc) · 6.5 KB
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package Algorithms.algorithm.dp;
import java.util.ArrayList;
public class MinAdjustmentCost_Class {
public static void main(String[] strs) {
ArrayList<Integer> A = new ArrayList<Integer>();
A.add(1);
A.add(4);
A.add(2);
A.add(3);
System.out.println(MinAdjustmentCost1(A, 1));
}
/**
* @param A: An integer array.
* @param target: An integer.
*/
public static int MinAdjustmentCost1(ArrayList<Integer> A, int target) {
// write your code here
if (A == null) {
return 0;
}
ArrayList<Integer> B = new ArrayList<Integer>();
for (int i = 0; i < A.size(); i++) {
B.add(0);
}
return rec(A, B, target, 0);
}
/*
* SOL 1:
* 最普通的递归方法。
* */
public static int rec(ArrayList<Integer> A, ArrayList<Integer> B, int target, int index) {
int len = A.size();
if (index >= len) {
// The index is out of range.
return 0;
}
int dif = 0;
int min = Integer.MAX_VALUE;
// If this is the first element, it can be from 1 to 100;
for (int i = 0; i <= 100; i++) {
if (index != 0 && Math.abs(i - B.get(index - 1)) > target) {
continue;
}
B.set(index, i);
dif = Math.abs(i - A.get(index));
dif += rec(A, B, target, index + 1);
min = Math.min(min, dif);
// 回溯
//B.set(index, A.get(index));
}
return min;
}
/*
* 递归2：
* Rec + memory.
* */
/**
* @param A: An integer array.
* @param target: An integer.
*/
public static int MinAdjustmentCost2(ArrayList<Integer> A, int target) {
// write your code here
if (A == null || A.size() == 0) {
return 0;
}
int[][] M = new int[A.size()][100];
for (int i = 0; i < A.size(); i++) {
for (int j = 0; j < 100; j++) {
M[i][j] = Integer.MAX_VALUE;
}
}
return rec2(A, new ArrayList<Integer>(A), target, 0, M);
}
public static int rec2(ArrayList<Integer> A, ArrayList<Integer> B, int target, int index,
int[][] M) {
int len = A.size();
if (index >= len) {
// The index is out of range.
return 0;
}
int dif = 0;
int min = Integer.MAX_VALUE;
// If this is the first element, it can be from 1 to 100;
for (int i = 1; i <= 100; i++) {
if (index != 0 && Math.abs(i - B.get(index - 1)) > target) {
continue;
}
if (M[index][i - 1] != Integer.MAX_VALUE) {
dif = M[index][i - 1];
min = Math.min(min, dif);
continue;
}
B.set(index, i);
dif = Math.abs(i - A.get(index));
dif += rec2(A, B, target, index + 1, M);
min = Math.min(min, dif);
// Record the result.
M[index][i - 1] = dif;
// 回溯
B.set(index, A.get(index));
}
return min;
}
/*
* SOLUTION 3 递归2：
* Rec + memory.
* 改进的递归版本
* */
/**
* @param A: An integer array.
* @param target: An integer.
*/
public static int MinAdjustmentCost3(ArrayList<Integer> A, int target) {
// write your code here
if (A == null || A.size() == 0) {
return 0;
}
int[][] M = new int[A.size()][100];
for (int i = 0; i < A.size(); i++) {
for (int j = 0; j < 100; j++) {
M[i][j] = Integer.MAX_VALUE;
}
}
// 首个数字可以取1-100
int min = Integer.MAX_VALUE;
for (int i = 1; i <= 100; i++) {
min = Math.min(min, rec3(A, target, 0, i, M));
}
return min;
}
/*
* 将当前值设置为x能求得的最小解
* */
public static int rec3(ArrayList<Integer> A, int target, int index, int x,
int[][] M) {
int len = A.size();
if (index >= len) {
// The index is out of range.
return 0;
}
if (M[index][x - 1] != Integer.MAX_VALUE) {
return M[index][x - 1];
}
int bas = Math.abs(x - A.get(index));
int min = Integer.MAX_VALUE;
// 对下一个值尝试取1-100
for (int i = 1; i <= 100; i++) {
// 下一个值的取值不可以超过abs
if (index != len - 1 && Math.abs(i - x) > target) {
continue;
}
// 计算dif
int dif = bas + rec3(A, target, index + 1, i, M);
min = Math.min(min, dif);
}
// Record the result.
M[index][x - 1] = min;
return min;
}
/*
* SOLUTION 4：
* DP
* */
/**
* @param A: An integer array.
* @param target: An integer.
*/
public static int MinAdjustmentCost(ArrayList<Integer> A, int target) {
// write your code here
if (A == null || A.size() == 0) {
return 0;
}
// D[i][v]: 把index = i的值修改为v，所需要的最小花费
int[][] D = new int[A.size()][101];
int size = A.size();
for (int i = 0; i < size; i++) {
for (int j = 1; j <= 100; j++) {
D[i][j] = Integer.MAX_VALUE;
if (i == 0) {
// The first element.
D[i][j] = Math.abs(j - A.get(i));
} else {
for (int k = 1; k <= 100; k++) {
// 不符合条件
if (Math.abs(j - k) > target) {
continue;
}
int dif = Math.abs(j - A.get(i)) + D[i - 1][k];
D[i][j] = Math.min(D[i][j], dif);
}
}
}
}
int ret = Integer.MAX_VALUE;
for (int i = 1; i <= 100; i++) {
ret = Math.min(ret, D[size - 1][i]);
}
return ret;
}
}