657. Robot Return to Origin
题目 657. Robot Return to Origin
思路分析
class Solution {
public boolean judgeCircle(String moves) {
StringBuilder stack = new StringBuilder();
for(int i=0;i<moves.length();i++){
char cur = moves.charAt(i);
if(stack.length()>0){
char top=stack.charAt(stack.length()-1);
if(isPair(top,cur)){
stack.deleteCharAt(stack.length()-1);
continue;
}
}
stack.append(cur);
}
return stack.length()==0;
}
private boolean isPair(char a,char b){
if (a == 'U' && b == 'D') return true;
if (a == 'D' && b == 'U') return true;
if (a == 'L' && b == 'R') return true;
if (a == 'R' && b == 'L') return true;
return false;
}
}
但其实发现会错 聪明反被聪明误了
栈的一个核心特性是必须消除相邻（或经消除后相邻）的元素。但这道题是二维平面的移动，X轴 的移动和 Y轴 的移动是互不干扰的，中间隔着别的方向也能抵消。
其实只需要看数量++ –-最后等不等于0即可
代码实现
class Solution {
public boolean judgeCircle(String moves) {
int[] cnt = new int[26];
for(char c:moves.toCharArray()){
cnt[c-'A']++;
}
return cnt['U'-'A']==cnt['D'-'A'] &&
cnt['L'-'A']==cnt['R'-'A'];
}
}
class Solution {
public boolean judgeCircle(String moves) {
int x = 0, y = 0;
for (int i = 0; i < moves.length(); i++) {
switch (moves.charAt(i)) {
case 'U': y++; break;
case 'D': y--; break;
case 'L': x--; break;
case 'R': x++; break;
}
}
return x == 0 && y == 0;
}
}
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