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notes/剑指 offer 题解.md
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@ -101,23 +101,36 @@
这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。 这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。
以 (2, 3, 1, 0, 2, 5) 为例，以下代码的运行过程为：
```html
position-0 : (2,3,1,0,2,5) // 2 <-> 1
(1,3,2,0,2,5) // 1 <-> 2
(3,1,1,0,2,5) // 3 <-> 0
(0,1,1,3,2,5) // already in position
position-1 : (0,1,1,3,2,5) // already in position
position-2 : (0,1,1,3,2,5) // nums[i] == nums[nums[i]], exit
```
遍历到位置 2 时，该位置上的数为 1，但是第 1 个位置上已经有一个 1 的值了，因此可以知道 1 重复。
```java ```java
public boolean duplicate(int[] numbers, int length, int[] duplication) { public boolean duplicate(int[] nums, int length, int[] duplication) {
if(numbers == null || length <= 0) return false; if (nums == null || length <= 0) return false;
for (int i = 0; i < length; i++) { for (int i = 0; i < length; i++) {
while (numbers[i] != i && numbers[i] != numbers[numbers[i]]) { while (nums[i] != i && nums[i] != nums[nums[i]]) {
swap(numbers, i, numbers[i]); swap(nums, i, nums[i]);
} }
if (numbers[i] != i && numbers[i] == numbers[numbers[i]]) { if (nums[i] != i && nums[i] == nums[nums[i]]) {
duplication[0] = numbers[i]; duplication[0] = nums[i];
return true; return true;
} }
} }
return false; return false;
} }
private void swap(int[] numbers, int i, int j) { private void swap(int[] nums, int i, int j) {
int t = numbers[i]; numbers[i] = numbers[j]; numbers[j] = t; int t = nums[i]; nums[i] = nums[j]; nums[j] = t;
} }
``` ```