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LeetCode题解（java语言实现） PDF 下载
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1.3 Solution 3 - Reversal
Can we do this in O(1) space and in O(n) time? The following solution does.
Assuming we are given 1,2,3,4,5,6 and order 2. The basic idea is:
1. Divide the array two parts: 1,2,3,4 and 5, 6
2. Rotate first part: 4,3,2,1,5,6
3. Rotate second part: 4,3,2,1,6,5
4. Rotate the whole array: 5,6,1,2,3,4
public static void rotate(int[] arr, int order) {
order = order % arr.length;
if (arr == null || order < 0) {
throw new IllegalArgumentException("Illegal argument!");
}
//length of first part
int a = arr.length - order;
reverse(arr, 0, a-1);
reverse(arr, a, arr.length-1);
reverse(arr, 0, arr.length-1);
}
public static void reverse(int[] arr, int left, int right){
if(arr == null || arr.length == 1)
return;
while(left < right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;
8 | 181 Program Creek
} }2 Evaluate Reverse Polish Notation
The problem:
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another
expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
2.1 Naive Approach
This problem is simple. After understanding the problem, we should quickly realize
that this problem can be solved by using a stack. We can loop through each element
in the given array. When it is a number, push it to the stack. When it is an operator,
pop two numbers from the stack, do the calculation, and push back the result.
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