Pascal's Triangle Leetcode Java and C++

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,
Return

[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]
]

太久没刷题觉得自己啥也不会了。。。但其实不要有畏难情绪是最重要的，做起来就发现还蛮简单的。。。

public class Solution {public List<List<Integer>> generate(int numRows) {List<List<Integer>> result = new ArrayList<>();if (numRows <= 0) {return result;}for (int i = 0; i < numRows; i++) {List<Integer> array = new ArrayList<>();array.add(1);if (i > 0) {List<Integer> pre = result.get(i - 1);int size = pre.size();for (int j = 0; j < size; j++) {if (j == size - 1) {array.add(pre.get(j));} else {array.add(pre.get(j) + pre.get(j + 1));}}}result.add(array);}return result;}
}

看了top solution后的改进版本，减少了条件判断：

public class Solution {public List<List<Integer>> generate(int numRows) {List<List<Integer>> result = new ArrayList<>();if (numRows <= 0) {return result;}for (int i = 0; i < numRows; i++) {List<Integer> array = new ArrayList<>();for (int j = 0; j < i + 1; j++) {if (j == 0 || j == i) {array.add(1);} else {List<Integer> pre = result.get(i - 1);array.add(pre.get(j - 1) + pre.get(j));}}result.add(array);}return result;}
}

但看了另一个top solution还是觉得可能自己就是写不好代码了。。。= =？

但也许我写的比较快一点吧？并不好判断。。。

public class Solution {public List<List<Integer>> generate(int numRows) {List<List<Integer>> result = new ArrayList<>();List<Integer> array = new ArrayList<>();if (numRows <= 0) {return result;}for (int i = 0; i < numRows; i++) {array.add(0, 1);for (int j = 1; j < array.size() - 1; j++) {array.set(j, array.get(j) + array.get(j + 1));}result.add(new ArrayList<>(array));}return result;}
}

附上c++的解法，和第一种解法的改进版是一样的：

class Solution {
public:vector<vector<int>> generate(int numRows) {vector<vector<int>> r(numRows);for (int i = 0; i < numRows; i++) {r[i].resize(i + 1);r[i][0] = 1, r[i][i] = 1;for (int j = 1; j < i; j++) {r[i][j] = r[i - 1][j - 1] + r[i - 1][j];}}return r;}
};

这么一看c++还真的挺简洁的。

转载于:https://www.cnblogs.com/aprilyang/p/6943158.html

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