"Leetcode-Interview algorithm classic-java implementation" "059-spiral Matrix II (Spiral Matrix II)"

"059-spiral Matrix II (Spiral Matrix II)" "leetcode-Interview algorithm classic-java Implementation" "All Topics folder Index" Original Question

Given an integer n, generate a square matrix filled with elements from 1 to N2 in Spiral order.
For example,
Given n = 3,
You should return the following matrix:

123894765 ]]

Main Topic

Given an integer n. Generates a n*n matrix, which is filled with 1-n^2 numbers.

Thinking of solving problems

The calculation method is used to calculate the corresponding number for each position.

Code Implementation

Algorithm implementation class

 Public  class solution {     Public int[][]Generatematrix(intN) {int[[] result =New intN [n];intLayerintK for(inti =0; I < n; i++) { for(intj =0; J < N; J + +) {layer = Layer (I, J, N);///current coordinates are several levels away                //n * n-layer * The last digit (also the largest) used by the outer layer of the layer                the first number used by the current layer where the coordinates are locatedk = n * N-(N-2* Layer) * (N-2* Layer) +1; RESULT[I][J] = k;//(N-2 * layer-1): Four (N-2 * layer-1) is the total number of elements (x, y) in the same layer                if(i = = layer) {//Case One, the coordinates are near the upper boundaryRESULT[I][J] = k + J-layer; }Else if(j = = N-layer-1) {//Situation two, coordinates off the right edge of the recentRESULT[I][J] = k + (N-2* Layer-1) + I-layer; }Else if(i = = N-layer-1) {//Case three, the coordinates are near the lower boundaryRESULT[I][J] = k +3* (N-2* Layer-1)-(J-layer); }Else{//Case three, the coordinates are near the left edgeRESULT[I][J] = k +4* (N-2* Layer-1)-(I-layer); }            }        }returnResult }/** * In a n*n matrix, calculate the number of layers outside the (x, y) coordinates, the coordinates start at 0 * * @param x Horizontal axis * @param y ordinate * @ param n Matrix size * Number of layers outside @return coordinates */     Public int Layer(intXintYintN) {x = x < n-1-X?
 x : n - 1 - x; // 计算横坐标离上下边界的近期距离 y = y < n - 1 - y ? y : n - 1 - y; // 计算纵坐标离左右边界的近期距离 return x < y ? x : y; // 较小的值为坐标的外围层数 }}

Assessment Results

　　Click on the picture, the mouse does not release, drag a position, release after the new form to view the full picture.

Special Instructions Welcome reprint. Reprint Please specify the source "http://blog.csdn.net/derrantcm/article/details/47164439"

"Leetcode-Interview algorithm classic-java implementation" "059-spiral Matrix II (Spiral Matrix II)"