题目描述
https://leetcode.com/problems/matrix-diagonal-sum/
博主提交的代码
class Solution {
public int diagonalSum(int[][] c) {
int currentOffSet = -1;
int result = 0;
for(int i = 0; i < c.length; i++){
if( isOdd( c.length) && i == c.length / 2 ){
result+=c[i][i];
} else{
result+=c[i][i];
result+=c[c.length-i - 1][i];
}
}
return result;
}
public boolean isOdd(int input){
if ( (input & 1) == 1){
return true;
} else {
return false;
}
}
}
他人优秀的代码
https://leetcode.com/problems/matrix-diagonal-sum/discuss/830726/Java-O(n)
我个人不喜欢对于重复相加中心点后，再做减法的解法，那样理解不直观，提前判断不重复加更加直观一些
public int diagonalSum(int[][] mat) {
int n = mat.length, res = 0;
for (int i = 0; i < n; i++) {
res += mat[i][i];
if (i != n - 1 - i) res += mat[i][n - 1 - i];
}
return res;
}
