小编CK9*_*920的帖子

我有一些包含嵌套if语句的代码:

if(numberOfNeighbors == 1){

                //go through comparison again, add Pixel(i,j) to current linked list -> complist[numberOfComponents]
                    //  break out of large check ??

                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j-1))){ //compare to top left
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break; 
                    }
                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i,j-1))){ // compare to top
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }

                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i+1,j-1))){ // compare to top right
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }
                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j))){ // compare to left
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }
} // end of if(numberOfNeighbors == 1)

基本上我想做的事情,无论效率如何低,都是比较一下4次,但如果事实证明它是一个匹配,则突破4个嵌套if语句的集合,以及外部if语句. …