标签：arraylist 背包 问题 ima 个数 背包问题 should not adjust
Given an integer array,
adjust each integers so that the difference of every adjacent integers are not greater than a given number target. If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]| Notice You can assume each number in the array is a positive integer and not greater than 100. Have you met this question in a real interview? Yes Example Given [1,4,2,3] and target = 1, one of the solutions is [2,3,2,3], the adjustment cost is 2 and it‘s minimal. Return 2.
这道题要看出是背包问题，不容易，跟FB一面 paint house很像，比那个难一点
定义res[i][j] 表示前 i个number with 最后一个number是j，这样的minimum adjusting cost
public int MinAdjustmentCost(ArrayList<Integer> A, int target) {
// write your code here
// 前i-1 个数调整后并且第i-1个数调整为j的cost
int n = A.size();
int[][] f = new int[n + 1][101];
// initialize
for (int i = 0; i <= 100; i++) {
f[0][i] = 0;
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= 100; j++) {
f[i][j] = Integer.MAX_VALUE;
}
}
//
//function
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= 100; j++) {
for (int k = 0; k <= 100; k++) {
if (Math.abs(j - k) <= target) {
f[i][k] = Math.min(f[i][k], f[i - 1][j] + Math.abs(A.get(i - 1) - k));
}
}
}
}
int ans = Integer.MAX_VALUE;
for (int i = 0; i <= 100; i++) {
ans = Math.min(ans, f[n][i]);
}
return ans;
}
标签：arraylist 背包 问题 ima 个数 背包问题 should not adjust
原文地址：http://www.cnblogs.com/apanda009/p/7294627.html
