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The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G Y I R And then read line by line: “PAHNAPLSIIGYIR”Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example :Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI” Explanation: P I N A L S I G Y A H R P I
找到了周期 ( n u m R o w s ∗ 2 − 2 ) (numRows*2 - 2) (numRows∗2−2),头行表示为 k ∗ ( n u m R o w s ∗ 2 − 2 ) k*(numRows*2 - 2) k∗(numRows∗2−2)和尾行为 k ∗ ( n u m R o w s ∗ 2 − 2 ) + ( n u m R o w s − 1 ) k*(numRows*2 - 2) + (numRows - 1) k∗(numRows∗2−2)+(numRows−1)。但在斜边的规律不太一样,作罢。
leetcode solution1: sort by row
用curRow和goingDown两个游标来控制。当curRow为第一行或最后一行时,goingDown反向,由此直接按照Z顺序将s分行存入list中,最后合并成一行。class Solution { public String convert(String s, int numRows) { if (numRows == 1) return s; Listrows = new ArrayList<>(); for (int i = 0; i < Math.min(numRows, s.length()); i++) rows.add(new StringBuilder()); int curRow = 0; boolean goingDown = false; for (char c : s.toCharArray()) { rows.get(curRow).append(c); if (curRow == 0 || curRow == numRows - 1) goingDown = !goingDown; curRow += goingDown ? 1 : -1; } StringBuilder ret = new StringBuilder(); for (StringBuilder row : rows) ret.append(row); return ret.toString(); }}
leetcode solution2: visit by row
周期公式的解法,原来中间重叠的部分可以用两个公式来表达, i i i为当前行 k ∗ ( n u m R o w s ∗ 2 − 2 ) + i k*(numRows*2 - 2) + i k∗(numRows∗2−2)+i 和 ( k + 1 ) ∗ ( n u m R o w s ∗ 2 − 2 ) − i (k+1)*(numRows*2 - 2) - i (k+1)∗(numRows∗2−2)−i 这样整个算法可以写为两个公式: k ∗ ( n u m R o w s ∗ 2 − 2 ) + i k*(numRows*2 - 2) + i k∗(numRows∗2−2)+i ( k + 1 ) ∗ ( n u m R o w s ∗ 2 − 2 ) − i (k+1)*(numRows*2 - 2) - i (k+1)∗(numRows∗2−2)−i //表示斜边上的元素 虽说空间复杂度更低,但是没有solution 1更加便于理解class Solution { public String convert(String s, int numRows) { if (numRows == 1) return s; StringBuilder ret = new StringBuilder(); int n = s.length(); int cycleLen = 2 * numRows - 2; for (int i = 0; i < numRows; i++) { for (int j = 0; j + i < n; j += cycleLen) { ret.append(s.charAt(j + i)); if (i != 0 && i != numRows - 1 && j + cycleLen - i < n) ret.append(s.charAt(j + cycleLen - i)); } } return ret.toString(); }}
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