354. Russian Doll Envelopes

DP(TLE)

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class Solution {
public int maxEnvelopes(int[][] envelopes) {
Arrays.sort(envelopes, (o1, o2) -> {
int diff = o1[0] - o2[0];
if (diff == 0) {
return o2[1] - o1[1]; // 宽度相同时高度降序可以在计算时避免比较宽度,只需比较高度即可
} else {
return diff;
}
});

int n = envelopes.length;
int res = 0;
// 定义 dp[i] 为 envelopes[i] 能容纳下多少个信封
int[] dp = new int[n];

for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (envelopes[i][1] > envelopes[j][1]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
res = Math.max(res, dp[i]);
}

return res + 1;
}
}

DP

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class Solution {
public int maxEnvelopes(int[][] envelopes) {
Arrays.sort(envelopes, (o1, o2) -> {
int diff = o1[0] - o2[0];
if (diff == 0) {
return o2[1] - o1[1]; // 注意宽度相同时根据高度降序以避免将宽度相同的信封装入
} else {
return diff;
}
});

int n = envelopes.length;
int[] tails = new int[n];
int nextIndex = 0;
for (int[] envelope : envelopes) {
int height = envelope[1];
int insertIndex = findInsertIndex(tails, 0, nextIndex - 1, height);
tails[insertIndex] = height;
if (insertIndex == nextIndex) {
nextIndex++;
}
}

return nextIndex;
}

private int findInsertIndex(int[] tails, int left, int right, int target) {
while (left <= right) {
int mid = (left + right) >>> 1;
if (tails[mid] < target) {
left = mid + 1;
} else if (tails[mid] > target) {
right = mid - 1;
} else {
return mid;
}
}

// now: [right, left]
return left;
}
}

难点在于对高度进行逆序排序并转换为 LIS 问题。

References

354. Russian Doll Envelopes