假设我有一个int [size] [size]矩阵.我想像这样遍历它:
01 02 03 04
08 07 06 05
09 10 11 12
16 15 14 13
我真的必须说:
for (int y=0; y<size; y++) {
int startIndex, endIndex, increment;
boolean lessThan;
if (y%2 == 0) {
startIndex = 0;
endIndex = size;
lessThan = true;
increment = 1;
}
else {
startIndex = size-1;
endIndex = -1;
lessThan = false;
increment = -1;
}
for (int x=startIndex; lessThan ? x<endIndex : x>endIndex; x+=increment) {
// Process element
}
}
为了避免这样做:
for (int y=0; y<size; y++) {
if (y%2 == 0) {
for (int x=0; x<size; x++) {
// Process element
}
}
else {
for (int x=size-1; x>-1; x--) {
// Process element
}
}
}
更好,但是重复了代码.有没有更好的方法来构造它?
解决方法:
遍历矩阵的另一种方法:
for (int num = 0; num < maxX*maxY; ++num) { // num - element number
// calc element coordinates
int y = num / maxX;
int x = num % maxX;
if (y % 2 == 1)
x = maxX - x;
int elem = matrix[x][y];
// process element
}