LC.P1713[得到子序列的最少操作次数]

方法一:贪心+动态规划+二分

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class Solution {
public int minOperations(int[] target, int[] arr) {
int n = target.length;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < n; ++i) map.put(target[i], i);
List<Integer> list = new ArrayList<>();
for (int x : arr) {
if (map.containsKey(x)) {
list.add(map.get(x));
}
}
int length = list.size(), max = 0;
int[] f = new int[length + 1], g = new int[length + 1];
Arrays.fill(g, Integer.MAX_VALUE);
for (int i = 0; i < length; ++i) {
int index = list.get(i);
int left = 0, right = length;
while (left < right) {
int mid = left + right + 1 >> 1;
if (g[mid] < index) left = mid;
else right = mid - 1;
}
int t = right + 1;
f[i] = t;
g[t] = Math.min(g[t], index);
max = Math.max(max, f[i]);
}
return n - max;
}
}
  • 时间复杂度:$O(n + mlogm)$
  • 空间复杂度:$O(n + m)$