题目如下
We'll say that a String is xy-balanced if for all the 'x' chars in the string, there exists a 'y' char somewhere later in the string. So "xxy" is balanced, but "xyx" is not. One 'y' can balance multiple 'x's. Return true if the given string is xy-balanced.
xyBalance("aaxbby") → true
xyBalance("aaxbb") → false
xyBalance("yaaxbb") → false
我的代码
/**
此题可简化为判别y是否比x更靠近右，巧妙运用indexOf()方法中，不存在则返回-1的特性实现如下代码：
*/
public boolean xyBalance(String str) {
int xIndex = str.lastIndexOf("x");//找到最右边的x
int yIndex = str.lastIndexOf("y");//找到最右边的y
return yIndex>=xIndex;//判断y是否在x的右边。
}
官方solution
public boolean xyBalance(String str) {
// Find the rightmost y
int y = -1;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i)=='y') y = i;
}
// Look at the x's, return false if one is after y
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i)=='x' && i > y) return false;
}
return true;
// Solution notes: this solution uses two loops, each written the simple
// 0..length way. You could do it with a single reverse loop, noticing
// if you see an x before a y. Or use lastIndexOf().
}
