[codility]Min-abs-sum
https://codility.com/demo/take-sample-test/delta2011/
0-1背包问题的应用。我自己一开始没想出来。“首先对数组做处理，负数转换成对应正数，零去掉，计数排序统计有多少个不同元素及其对应个数，并累加所有数的和sum，不妨记b=sum/2，不同元素个数为m，则目标转换为在m个不同元素中挑出若干个元素（每个元素可以重复多次，但少于它们的出现次数），使得它们的和不大于b并尽量接近。到了这里，应该有点熟悉的感觉了吧。对了，其实这就是0-1背包问题！” 参考http://phiphy618.blogspot.jp/2013/05/codility-delta-2011-minabssum.html
第一次的代码并未完全通过，75分，大数据全挂。原因是这里一个元素可以出现多次，是多重背包问题。
// you can also use imports, for example:
// import java.math.*;
class Solution {
public int solution(int[] A) {
// write your code here...
if (A.length == 0) return 0;
int sum = 0;
int max = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] < 0) A[i] = -A[i];
sum += A[i];
}
int target = sum / 2;
int dp[][] = new int[A.length][target];
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < target; j++) {
// j+1 is the weight limit
if (i == 0)
{
if (A[i] <= (j+1)) {
dp[i][j] = A[i];
}
else
{
dp[i][j] = 0;
}
}
else // i != 0
{
int w1 = dp[i-1][j];
int w2 = 0;
if (j-A[i] >=0 ) {
w2 = dp[i][j-A[i]] + A[i];
}
dp[i][j] = w1 > w2 ? w1 : w2;
}
}
}
max = dp[A.length - 1][target - 1];
return (sum - max * 2);
}
}
第二次参考了cp博士的文章，处理了多重背包的优化，并用了滚动数组：http://blog.csdn.net/caopengcs/article/details/10028269
// you can also use includes, for example:
// #include <algorithm>
int solution(const vector<int> &A) {
// write your code in C++98
int len = A.size();
int sum = 0;
int M = 0;
for (int i = 0; i < len; i++) {
int x = 0;
x = A[i] > 0 ? A[i] : -A[i];
sum += x;
if (x > M)
M = x;
}
vector<int> count;
count.resize(M+1);
for (int i = 0; i < len; i++) {
int x = 0;
x = A[i] > 0 ? A[i] : - A[i];
count[x]++;
}
int target = sum / 2;
int largest = 0;
vector<int> dp(target+1, -1);
for (int i = 0; i <= M; i++) {
if (count[i] > 0) {
for (int j = 0; j <= target; j++) {
if (j == 0) dp[j] = count[i];
if (dp[j] >= 0) {
dp[j] = count[i];
if (j > largest)
largest = j;
}
else if (j - i >= 0 && dp[j - i] > 0) {
dp[j] = dp[j - i] - 1;
if (j > largest)
largest = j;
}
else {
dp[j] = -1;
}
}
}
}
return abs(sum - 2 * largest);
}