LC.P354[俄罗斯套娃信封问题]

方法一：动态规划

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class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        int n = envelopes.length;
        if (n == 0) return 0;

        Arrays.sort(envelopes, (a, b) -> {
            if (a[0] != b[0]) return a[0] - b[0]; // 按宽度升序
            else return b[1] - a[1]; // 按高度降序
        });

        int[] f = new int[n];
        Arrays.fill(f, 1);
        int ans = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (envelopes[j][1] < envelopes[i][1]) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
            ans = Math.max(ans, f[i]);
        }
        return ans;
    }
}
  • 时间复杂度：$O(n^2)$
  • 空间复杂度：$O(n)$

方法二：二分查找+动态规划

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class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        int n = envelopes.length;
        if (n == 0) return 0;
        Arrays.sort(envelopes, (a, b) -> {
            if (a[0] != b[0]) return a[0] - b[0]; // 按宽度升序
            else return b[1] - a[1]; // 按高度降序
        });
        List<Integer> list = new ArrayList<>();
        list.add(envelopes[0][1]);
        for (int i = 1; i < n; ++i) {
            int h = envelopes[i][1];
            if (h > list.get(list.size() - 1)) {
                list.add(h);
            } else {
                int index = binarySearch(list, h);
                list.set(index, h);
            }
        }
        return list.size();
    }

    private int binarySearch(List<Integer> list, int target) {
        int left = 0, right = list.size() - 1;
        while (left < right) {
            int mid = left + right >> 1;
            if (list.get(mid) >= target) right = mid;
            else left = mid + 1;
        }
        return left;
    }
}
  • 时间复杂度：$O(nlogn)$
  • 空间复杂度：$O(n)$