思路
操作次数0-nums1.size()之间
时间复杂度： O(n²)
空间复杂度： O(n)
Code：
class Solution {
public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {
int n = nums1.size();
int[][] f = new int[n + 1][n + 1];
int[][] nums = new int[n][0];
for (int i = 0; i < n; ++i) {
nums[i] = new int[] {
nums1.get(i), nums2.get(i)};
}
Arrays.sort(nums, Comparator.comparingInt(a -> a[1]));
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j > 0) {
int a = nums[i - 1][0], b = nums[i - 1][1];
f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + a + b * j);
}
}
}
int s1 = 0, s2 = 0;
for (int v : nums1) {
s1 += v;
}
for (int v : nums2) {
s2 += v;
}
for (int j = 0; j <= n; ++j) {
if (s1 + s2 * j - f[n][j] <= x) {
return j;
}
}
return -1;
}
}
注:不会，参考了题解。。。。。。
原文链接: https://blog.csdn.net/qq_53568730/article/details/135693563