1138. 字母板上的路径
1.想法
image.png从a出发到任意的点例如 l,最直接的就是先走竖行再走横行
image.png
有一个特殊情况是Z,只能向上走
所以,我们的步骤是.先得到当前的位置信息,和要去的位置信息,最后进行移动
2:代码
class Solution {
public String alphabetBoardPath(String target) {
StringBuilder sb = new StringBuilder();
char[][] board = new char[6][5];
board[0] = new char[]{'a','b','c','d','e'};
board[1] = new char[]{'f','g','h','i','j'};
board[2] = new char[]{'k','l','m','n','o'};
board[3] = new char[]{'p','q','r','s','t'};
board[4] = new char[]{'u','v','w','x','y'};
board[5] = new char[]{'z','0','0','0','0'};
//字母板
char[] chars = target.toCharArray();
int startX=0,startY=0;
char nowCh ='a';
for(char c:chars){
if(nowCh == c){ //如果当前的字母和上一个相同,直接添加!
sb.append("!");
}else {
int[] pos = getMyCharPos(board, c); //先得到位置
sb.append(getMyOrder(startX, startY, pos[0], pos[1]));//添加移动指令
startX = pos[0]; //移动到指令位置
startY = pos[1];
}
nowCh = c; //更新该char
}
return sb.toString();
}
private String getMyOrder(int startX, int startY, int endX, int endY) {
StringBuilder sb = new StringBuilder();
boolean flag = false;
if(startX ==5&&startY == 0){
sb.append("U");
startX--;
}
if(endX == 5&&endY==0){
endX--;
flag = true;
}
//处理X
if(startX<endX){
while(endX!=startX){
sb.append("D");
startX++;
}
}else{
while(endX!=startX){
sb.append("U");
startX--;
}
}
//处理Y
if(startY<endY){
while(endY!=startY){
sb.append("R");
startY++;
}
}else{
while(endY!=startY){
sb.append("L");
startY--;
}
}
if(flag){
sb.append("D");
}
sb.append("!");
return sb.toString();
}
private int[] getMyCharPos(char[][] board, char c) {
int[] pos = new int[2];
for(int i=0;i<board.length;i++){
if(board[i][0] == c){
pos[0] = i;
pos[1] = 0;
break;
}
if(board[i][0] <c&&board[i+1][0]>c){
for(int j=0;j<board[i].length;j++){
if(board[i][j] == c){
pos[0] = i;
pos[1] = j;
}
}
}
}
return pos;
}
}
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