[Leetcode]: 48:rotate Image

Topic:

You are given a n x n 2D matrix representing an image.

Rotate the image by degrees (clockwise).

Follow up:
Could do this in-place?

Train of thought 1: Find correspondence, 90 degree flip is: Old ordinate--new horizontal axis New ordinate = the old horizontal axis inversion (maximum height-the old horizontal axis)

     Public voidRotateint[] matrix) {        int[] Temp =New int[Matrix.length] [Matrix[0].length];  for(inti = 0; i< matrix.length;i++){             for(intj = 0; j< matrix.length;j++) {Temp[j][matrix.length-1-i] =Matrix[i][j]; }        }                 for(inti = 0; i< matrix.length;i++){             for(intj = 0; j< matrix.length;j++) {Matrix[i][j]=Temp[i][j]; }        }    }

Analysis: This method of problem-solving uses extra storage space!

Idea 2: Flip + fold

90 degree conversion = Flip (diagonal flip at any angle (apostrophe to flip): a[i][j] = a[j][i]) + fold (upside down or left/right flip): a[i][j] = a[i][length-1-j])

Note that 270 degrees is a similar idea.

Promotion: Flip 180 degrees = Flip (diagonal flip at any angle: a[i][j] = a[j][i])

Code:

     Public voidRotateint[] matrix) {                //to Flip         for(inti = 0; i< matrix.length;i++){             for(intj = 0; J<i; j + +){                intTemp =Matrix[i][j]; MATRIX[I][J]=Matrix[j][i]; Matrix[j][i]=Temp; }        }                //Vertical Rollover         for(inti = 0; i< matrix.length;i++){             for(intj = 0; j< matrix.length/2;j++){                intTemp = matrix[i][matrix.length-1-J]; Matrix[i][matrix.length-1-J] =Matrix[i][j]; MATRIX[I][J]=Temp; }        }    }

[Leetcode]: 48:rotate Image