92. Backpack
可行性
https://www.lintcode.com/problem/backpack/description?_from=ladder&&fromId=16
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
public int backPack(int m, int[] A) {
// write your code here
int n = A.length;
if(n==0){
return 0;
}
//前i个物品拼出j的重量
boolean[][] f = new boolean[n+1][m+1];
for(int i =1;i<=m;i++){
f[0][i]= false;
}
f[0][0]= true;
for(int i=1;i<=n;i++){
for(int w =0;w<=m;w++){
f[i][w] = f[i-1][w];
if(w>=A[i-1]){
f[i][w]=(f[i][w] || f[i-1][w-A[i-1]]);
}
}
}
for(int i = m;i>=0;i--){
if(f[n][i]){
return i;
}
}
return 0;
}
}
滚动数组优化版：
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
public int backPack(int m, int[] A) {
// write your code here
int n = A.length;
if(n==0){
return 0;
}
//前i个物品拼出j的重量
boolean[][] f = new boolean[n+1][m+1];
int old =0;
int now =0;
for(int i =1;i<=m;i++){
f[now][i]= false;
}
f[now][0]= true;
for(int i=1;i<=n;i++){
old = now;
now = 1-now;
for(int w =0;w<=m;w++){
f[now][w] = f[old][w];
if(w>=A[i-1]){
f[now][w]=(f[now][w] || f[old][w-A[i-1]]);
}
}
}
for(int i = m;i>=0;i--){
if(f[now][i]){
return i;
}
}
return 0;
}
}
正整数求和可考虑背包问题
563. Backpack V
计数型
https://www.lintcode.com/problem/backpack-v/?_from=ladder&&fromId=16
public class Solution {
/**
* @param nums: an integer array and all positive numbers
* @param target: An integer
* @return: An integer
*/
//f[i][w] = f[i-1][w]+f[i-1][w-A[j]]
//初始条件：f[0][0] = 1
// f[0][1...m] = 0;
public int backPackV(int[] nums, int target) {
// write your code here
if(nums==null || nums.length==0){
return 0;
}
int n = nums.length;
int[][] f = new int[n+1][target+1];
for(int i=1;i<=target;i++){
f[0][i] =0;
}
f[0][0] = 1;
for(int i =1;i<=n;i++){
for(int w =0;w<=target;w++){
f[i][w]= f[i-1][w];
if(w>=nums[i-1]){
f[i][w]+=f[i-1][w-nums[i-1]];
}
}
}
return f[n][target];
}
}
空间优化成一维滚动数组
public class Solution {
/**
* @param nums: an integer array and all positive numbers
* @param target: An integer
* @return: An integer
*/
//f[i][w] = f[i-1][w]+f[i-1][w-A[j]]
//初始条件：f[0][0] = 1
// f[0][1...m] = 0;
public int backPackV(int[] nums, int target) {
// write your code here
if(nums==null || nums.length==0){
return 0;
}
int n = nums.length;
int[] f = new int[target+1];
for(int i=1;i<=target;i++){
f[i] =0;
}
f[0] = 1;
for(int i =1;i<=n;i++){
//right -> left
for(int w =target;w>=nums[i-1];w--){
f[w]+=f[w-nums[i-1]]; //overwrit
}
}
return f[target];
}
}
564. Combination Sum IV
计数型
https://www.lintcode.com/problem/combination-sum-iv/description?_from=ladder&&fromId=16
public class Solution {
/**
* @param nums: an integer array and all positive numbers, no duplicates
* @param target: An integer
* @return: An integer
*/
public int backPackVI(int[] nums, int target) {
// write your code here
if(nums==null || nums.length==0){
return 0;
}
if(target==0){
return 0;
}
int[] f = new int[target+1];
f[0]=1;
for(int i =1;i<=target;i++){
for(int j=0;j<nums.length;j++){
if(i>=nums[j])
f[i]+=f[i-nums[j]];
}
}
return f[target];
}
}
125. Backpack II
最值型
https://www.lintcode.com/problem/backpack-ii/description?_from=ladder&&fromId=16
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @param V: Given n items with value V[i]
* @return: The maximum value
*/
public int backPackII(int m, int[] A, int[] V) {
// write your code here
if(A==null || A.length==0 || V==null || V.length ==0){
return 0;
}
int n = A.length;
int[][] f = new int[n+1][m+1];
f[0][0] = 0;
for(int i =1;i<=m;i++){
f[0][i] = -1;
}
for(int i=1;i<=n;i++){
for(int w=0;w<=m;w++){
f[i][w] = f[i-1][w];
if(w>=A[i-1] && f[i-1][w-A[i-1]]!=-1){
f[i][w] = Math.max(f[i][w],f[i-1][w-A[i-1]]+V[i-1]);
}
}
}
int res = 0;
for(int i =0;i<=m;i++){
res = Math.max(res,f[n][i]);
}
return res;
}
}
440. Backpack III
最值型
https://www.lintcode.com/problem/backpack-iii/description?_from=ladder&&fromId=16
思路：由Backpack II扩展出来，设k为i-1物体取用次数
public class Solution {
/**
* @param A: an integer array
* @param V: an integer array
* @param m: An integer
* @return: an array
*/
public int backPackIII(int[] A, int[] V, int m) {
// write your code here
if(A==null || A.length==0 || V==null || V.length==0){
return 0;
}
int n = A.length;
int[][]f = new int[n+1][m+1];
f[0][0] = 0;
for(int i=1;i<=m;i++){
f[0][i]=-1;
}
for(int i=1;i<=n;i++){
for(int w=0;w<=m;w++){
f[i][w] = f[i-1][w];
int currW = w;
int k =1;
while(currW>=A[i-1]){
if(f[i-1][currW-A[i-1]]!=-1){
f[i][w] = Math.max(f[i][w],f[i-1][currW-A[i-1]]+V[i-1]*k);
}
currW-=A[i-1];
k++;
}
}
}
int res = 0;
for(int i=1;i<=m;i++){
res = Math.max(res,f[n][i]);
}
return res;
}
}
优化1：观察公式，进行简化
优化思路，其实最后一个无论取几次，都相当于k-1次取值f[i][w-A[i-1]]+V[i-1];
public class Solution {
/**
* @param A: an integer array
* @param V: an integer array
* @param m: An integer
* @return: an array
*/
public int backPackIII(int[] A, int[] V, int m) {
// write your code here
if(A==null || A.length==0 || V==null || V.length==0){
return 0;
}
int n = A.length;
int[][]f = new int[n+1][m+1];
f[0][0] = 0;
for(int i=1;i<=m;i++){
f[0][i]=-1;
}
for(int i=1;i<=n;i++){
for(int w=0;w<=m;w++){
f[i][w] = f[i-1][w];
//优化思路，其实最后一个无论取几次，都相当于k-1次取值f[i][w-A[i-1]]+V[i-1];
if(w>=A[i-1] && f[i][w-A[i-1]]!=-1){
f[i][w] = Math.max(f[i][w],f[i][w-A[i-1]]+V[i-1]);
}
}
}
int res = 0;
for(int i=1;i<=m;i++){
res = Math.max(res,f[n][i]);
}
return res;
}
}
优化2:画图之后改成一维滚动数组
public class Solution {
/**
* @param A: an integer array
* @param V: an integer array
* @param m: An integer
* @return: an array
*/
public int backPackIII(int[] A, int[] V, int m) {
// write your code here
if(A==null || A.length==0 || V==null || V.length==0){
return 0;
}
int n = A.length;
int[]f = new int[m+1];
f[0] = 0;
for(int i=1;i<=m;i++){
f[i]=-1;
}
for(int i=1;i<=n;i++){
for(int w=0;w<=m;w++){
//优化思路，其实最后一个无论取几次，都相当于k-1次取值f[i][w-A[i-1]]+V[i-1];
if(w>=A[i-1] && f[w-A[i-1]]!=-1){
f[w] = Math.max(f[w],f[w-A[i-1]]+V[i-1]);
}
}
}
int res = 0;
for(int i=1;i<=m;i++){
res = Math.max(res,f[i]);
}
return res;
}
}
原文地址：https://www.cnblogs.com/lizzyluvcoding/p/10798998.html
时间： 2024-10-29 22:47:12