[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 limitif (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++98int 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);
}