354. 俄罗斯套娃信封问题

354. 俄罗斯套娃信封问题 - 力扣（LeetCode）

信封的宽高为envelope的二维数组
宽为升序，宽相等时，高为降序
envelopes[j][1] < envelopes[i][1] 为啥高维度比较的是1，不是0呢，，数学没学好啊。二维数组 n行2列
// envelopes = [[w, h], [w, h]...]   
f[i]表示以 envelopes[i][1] 这个数结尾的最长递增子序列的长度
class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        if (envelopes.length == 0) {
            return 0;
        }
        
        int n = envelopes.length;
        Arrays.sort(envelopes, new Comparator<int[]>() {
            public int compare(int[] e1, int[] e2) {
                if (e1[0] != e2[0]) {
                    return e1[0] - e2[0];
                } else {
                    return e2[1] - e1[1];
                }
            }
        });

        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;
    }
}

基于二分查找的动态规划

class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        if (envelopes.length == 0) {
            return 0;
        }
        
        int n = envelopes.length;
        Arrays.sort(envelopes, new Comparator<int[]>() {
            public int compare(int[] e1, int[] e2) {
                if (e1[0] != e2[0]) {
                    return e1[0] - e2[0];
                } else {
                    return e2[1] - e1[1];
                }
            }
        });

        List<Integer> f = new ArrayList<Integer>();
        f.add(envelopes[0][1]);
        for (int i = 1; i < n; ++i) {
            int num = envelopes[i][1];
            if (num > f.get(f.size() - 1)) {
                f.add(num);
            } else {
                int index = binarySearch(f, num);
                f.set(index, num);
            }
        }
        return f.size();
    }

    public int binarySearch(List<Integer> f, int target) {
        int low = 0, high = f.size() - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            if (f.get(mid) < target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
}