LeetCode Matrix Summary

题目：

1. Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].

【解题思路】

顺着spiralorder剥皮，up->right->bottom->left. 用四个变量x1, x2, y1, y2 控制上下左右边界。注意循环条件为x1<=x2 && y1<= y2

另外注意每行和每列输出的范围

vector<int> spiralOrder(vector<vector<int> > &matrix) {     vector<int> result;     if(matrix.size() == 0) return result;     int x1 = 0;     int y1 = 0;     int x2 = matrix.size() - 1;     int y2 = matrix[0].size() - 1;     while(x1<=x2 && y1<=y2){        //up row        for(int j = y1; j<=y2; ++j){            result.push_back(matrix[x1][j]);        }               // right column        for(int i = x1+1; i<=x2; ++i){            result.push_back(matrix[i][y2]);        }                if(x2 !=x1){        // bottom row        for(int j = y2-1; j>=y1; --j){            result.push_back(matrix[x2][j]);        }}               if(y2 != y1){        // left column        for(int i = x2-1; i>x1; --i){            result.push_back(matrix[i][y1]);        }}        x1++, y1++, x2--, y2--;     }     return result;}

2. 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:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]

【解题思路】

和第一题类似，反过来构造就可以了。Code可以复用

vector<vector<int> > generateMatrix(int n) {     vector<vector<int>> matrix(n);     if(n == 0) return matrix;     for(int i=0; i<n; i++){        matrix[i].resize(n);     }     int x1 = 0;     int y1 = 0;     int x2 = n - 1;     int y2 = n - 1;     int val = 1;     while(x1<=x2 && y1<=y2){        //up row        for(int j = y1; j<=y2; ++j){            matrix[x1][j] = val++;        }        // right column        for(int i = x1+1; i<=x2; ++i){            matrix[i][y2] = val++;        }        if(x2 !=x1){        // bottom row        for(int j = y2-1; j>=y1; --j){            matrix[x2][j] = val++;        }}        if(y2 != y1){        // left column        for(int i = x2-1; i>x1; --i){            matrix[i][y1] = val++;        }}        x1++, y1++, x2--, y2--;     }     return matrix;}

3. Set matrix zeros

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

如果矩阵中某一元素为零，则将其所在列与行置为零

click to show follow up.
需要对算法进行优化的点
Follow up:
Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

【解题思路】

解题点就在于清空标志位存在哪里。可以创建O(m+n)的数组来存储，选择第一行和第一列来存储标志位。

1.先确定第一行和第一列是否需要清零
2.扫描剩下的矩阵元素，如果遇到了0，就将对应的第一行和第一列上的元素赋值为0
3.根据第一行和第一列的信息，已经可以讲剩下的矩阵元素赋值为结果所需的值了
4.根据1中确定的状态，处理第一行和第一列。

void setZeroes(vector<vector<int> > &matrix) {    assert(matrix.size() >0);    int row = matrix.size();    int column = matrix[0].size();    bool zeroRow = false, zeroCol = false;    for(int i=0; i<column; i++){        if(matrix[0][i] == 0)            zeroRow = true;    }    for(int i=0; i<row; i++){        if(matrix[i][0] == 0)            zeroCol = true;    }    for(int i=1; i<row; i++){        for(int j=1; j<column; j++){            if(matrix[i][j]==0){                matrix[i][0] = 0;                matrix[0][j] = 0;            }        }    }    for(int i=1; i<row; i++){        for(int j=1; j<column; j++){            if(matrix[i][0]==0 || matrix[0][j]==0)                matrix[i][j] = 0;        }    }    if(zeroRow == true){        for(int i=0; i<column;i++){            matrix[0][i] = 0;        }    }    if(zeroCol == true){        for(int i=0; i<row;i++){            matrix[i][0] = 0;        }    }}

未完，待续...