LeetCode119 Pascal's Triangle II

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详细见：leetcode.com/problems/pascals-triangle-ii

Java Solution: github

package leetcode;import java.util.ArrayList;import java.util.List;/* * Given an index k, return the kth row of the Pascal's triangle.For example, given k = 3,Return [1,3,3,1].Note:Could you optimize your algorithm to use only O(k) extra space? */public class P119_PascalsTriangleII {public static void main(String[] args) {Solution s = new Solution();for (int i = -1; i < 11; i ++) {List<Integer> ans = s.getRow(i);tools.Utils.B_打印List_Integer_OneLine(ans);}}static class Solution {    public List<Integer> getRow(int rowIndex) {    rowIndex ++;    List<Integer> ans = new ArrayList<>(rowIndex > 0 ? rowIndex : 1);    if (rowIndex <= 1) {    ans.add(1);    return ans;    }    for (int i = 0; i < rowIndex; i ++) {    ans.add(0);    }    ans.set(rowIndex - 1, 1);    ans.set(rowIndex - 2, 1);    int index = rowIndex - 3;    while (index != - 1) {    for (int jndex = index + 1; jndex < rowIndex - 1; jndex ++) {    ans.set(jndex, ans.get(jndex) + ans.get(jndex + 1));    }    ans.set(index, 1);    index --;    }    return ans;    }}}

C Solution: github

/*    url: leetcode.com/problems/pascals-triangle-ii    AC 3ms 2.56%*/void swap(int** a, int** b) {    int* t = *a;    *a = *b;    *b = t;}int* getRow(int ri, int* rn) {    int* ans = (int*) malloc(sizeof(int) * (ri+1));    int* tmp = (int*) malloc(sizeof(int) * (ri+1));    int i = 0, j = 0, k = 0;    for (i = 0; i <= ri; i ++) {        ans[0] = 1;        for (j = 1; j <= i; j ++) {            ans[j] = tmp[j] + tmp[j-1];        }        ans[i] = 1;        swap(&ans, &tmp);    }    free(ans);    *rn = ri+1;    return tmp;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/pascals-triangle-ii    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年5月4日    @details:    Solution:  55ms 25.60%'''class Solution(object):    def getRow(self, i):        """        :type i: int        :rtype: List[int]        """        L, R = [1]*(i+1), [1]*(i+1)        if i < 2: return L        for k in range(2, i):            for j in range(1, k):                R[k][j] = R[k-1][j-1]+R[k-1][j]            L, R = R, L        return L

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