Rotate Matrix by One


http://www.cnblogs.com/yrbbest/p/5165084.html
记得有道Amazon的OA题目,好像是给定一个矩阵,让把矩阵的每个元素向右shift一个位置。这道题之前没有好好自习想过。今天正好刷到了rotate matrix,所以正好一块想了。 
思路是类似LeetCode Spiral Matrix:
  1. 假设矩阵为方阵
  2. 设置top, left, bot, right四个边界变量,然后从最外圈到最内圈一圈一圈的shift。
  3. 设定一个count,当count < total elements in matrix的时候进行shift
  4. 在每一圈开始的时候记录下来matrix[top][left],然后开始shift
    1. 从top到bot
    2. 从left到right
    3. 从bot到top
    4. 从right到left
      1. 在最后一条边的结果尝试更新新的matrix[top][left],这时候的top为旧的,而left已经更新过一次了, 我们要分为两种情况考虑
        1. count != totalElements - 1, 这时候我们要:
          1. matrix[top][left] 更新为tmp
          2. count++
          3. top++
          4. 进入下一圈
        2. 否则 count == totalElements - 1,也要分为两种情况
          1. totalElements为奇数,我们不改变matrix[top][left]
          2. totalElements为偶数,这时我们依然要更新一次matrix[top][left] = tmp
        3. 然后count++结束循环返回结果
  5. 假如给定矩阵不为方阵,则我们还要加入更多判断,比如剩下最后一行或者最后一列的时候不更新之类的。最后一行或者最后一列可以由bot - top 或者 right - left分别求出
Time Complexity - O(mn),Space Complexity - O(1),  in place。
    public void rotateMatrixByOne(int[][] matrix) {
        if (matrix == null || matrix[0] == null) {
            return;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        int left = 0, top = 0, right = n - 1, bot = m - 1;
        int count = 0;
        int totalElements = m * n;
        
        while (count < totalElements) {            
            int tmp = matrix[top][left];            
            if (count < totalElements) {
                for (int i = top; i < bot; i++) {
                    matrix[i][left] = matrix[i + 1][left];
                    count++;
                }
                left++;
            }
            if (count < totalElements) {
                for (int i = left - 1; i < right; i++) {
                    matrix[bot][i] = matrix[bot][i + 1];
                    count++;
                }
                bot--;
            }
            if (count < totalElements) {
                for (int i = bot + 1; i > top; i--) {
                    matrix[i][right] = matrix[i - 1][right];
                    count++;
                }
                right--;
            }
            if (count < totalElements) {
                for (int i = right + 1; i > left; i--) {
                    matrix[top][i] = matrix[top][i - 1];
                    count++;
                }
                if (count != totalElements - 1) {
                    matrix[top][left] = tmp;
                } else if (totalElements % 2 == 0) {
                    matrix[top][left] = tmp;
                }
                count++;
                top++;
            }            
        }        
    }

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