以这样的方式对给定字符串进行分区,即第i个子串是第(i-1)子串和第(i-2)子串的和

以这样的方式对给定字符串进行分区,即第i个子串是第(i-1)个和第(i-2)个子串的和。

null

例如:

Input : "11235813"
Output : ["1", "1", "2", "3", "5", "8", "13"]

Input : "1111223"
Output : ["1", "11", "12", "23"]

Input : "1111213"
Output : ["11", "1", "12", "13"]

Input : "11121114"
Output : []

1.从一个数字开始,一次选取三个数字(第一、第二和第三),遍历给定的字符串。 2.如果first+second=third,则递归调用check(),其中second为first,third为second。第三个是根据下一个可能的位数选择的。(两个数字相加的结果最多可包含第二位和第三位数字+1) 3.否则,首先增加第三个(通过增加更多数字)直到极限(这里极限是第一和第二的总和)。 4.增加三分之一后,出现以下情况。 a) 当不匹配时,增加第二个偏移量。 b) 当不匹配时,增加第一个偏移量。 c) 注意:在增加第三个偏移量后调用check()后,不要更改第二个或第一个偏移量,因为它们已经完成。 5.当到达字符串末尾且满足条件时,将“second”和“third”添加到空列表中。在回滚递归堆栈时,将“第一个”放在列表的前面,以便保留顺序。

// Java program to check if we can partition a
// string in a way that value of i-th string is
// sum of (i-1)-th and (i-2)-th substrings.
import java.util.LinkedList;
public class SumArray {
private static LinkedList<Integer> resultList =
new LinkedList<>();
private static boolean check( char [] chars, int offset1,
int offset2, int offset3, boolean freezeFirstAndSecond) {
// Find subarrays according to given offsets
int first = intOf(subArr(chars, 0 , offset1));
int second = intOf(subArr(chars, offset1, offset2));
int third = intOf(subArr(chars, offset1 + offset2, offset3));
// If condition is satisfied for current subarrays
if (first + second == third) {
// If whole array is covered.
if (offset1 + offset2 + offset3 >= chars.length) {
resultList.add(first);
resultList.add(second);
resultList.add(third);
return true ;
}
// Check if remaining array also satisfies the condition
boolean result = check(subArr(chars, offset1,
chars.length - offset1), offset2, offset3,
Math.max(offset2, offset3), true );
if (result) {
resultList.addFirst(first);
}
return result;
}
// If not satisfied, try incrementing third
if (isValidOffSet(offset1, offset2, 1 + offset3, chars.length)) {
if (check(chars, offset1, offset2, 1 + offset3, false ))
return true ;
}
// If first and second have been finalized, do not
// alter already computed results
if (freezeFirstAndSecond)
return false ;
// If first and second are not finalized
if (isValidOffSet(offset1, 1 + offset2, Math.max(offset1,
1 + offset2),  chars.length)) {
// Try incrementing second
if (check(chars, offset1, 1 + offset2,
Math.max(offset1, 1 + offset2), false ))
return true ;
}
// Try incrementing first
if (isValidOffSet( 1 + offset1, offset2, Math.max( 1 + offset1,
offset2),  chars.length)) {
if (check(chars, 1 + offset1, offset2, Math.max( 1 + offset1,
offset2), false ))
return true ;
}
return false ;
}
// Check if given three offsets are valid (Within array length
// and third offset can represent sum of first two)
private static boolean isValidOffSet( int offset1, int offset2,
int offset3, int length) {
return (offset1 + offset2 + offset3 <= length &&
(offset3 == Math.max(offset1, offset2) ||
offset3 == 1 + Math.max(offset1, offset2)));
}
// To get a subarray with starting from given
// index and offset
private static char [] subArr( char [] chars, int index, int offset) {
int trueOffset = Math.min(chars.length - index, offset);
char [] destArr = new char [trueOffset];
System.arraycopy(chars, index, destArr, 0 , trueOffset);
return destArr;
}
private static int intOf( char ... chars) {
return Integer.valueOf( new String(chars));
}
public static void main(String[] args) {
String numStr = "11235813" ;
char [] chars = numStr.toCharArray();
System.out.println(check(chars, 1 , 1 , 1 , false ));
System.out.println(resultList);
}
}


输出:

true
[1, 1, 2, 3, 5, 8, 13]

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