[leetcode]Next Permutation
2016-04-08 16:51
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思路:从数组的后面往前遍历,当找到一个位置 i 的数字比他的前一个位置i-1数字大之后。再从 i 位置向后找,找到一个刚好比i-1位置的数字大的数字j,交换这二个数字,然后从i位置到数组末尾,把数组翻转一遍就好了。
public class Solution {
public void nextPermutation(int[] nums) {
int len = nums.length;
boolean flag = false;
for(int i=len-1; i>=1; i--){
if(nums[i]>nums[i-1]){
int j=i;
while(j<len-1){
if(nums[j+1]<=nums[i-1] ){
break;
}
j++;
}
if(nums[j]>nums[i-1]){
swap(nums,i-1,j);
}
reverse(nums,i,len-1);
flag = true;
break;
}
}
if(!flag){
reverse(nums,0,len-1);
}
}
private void reverse(int[] nums,int i,int j){
while(i<j){
swap(nums,i++,j--);
}
}
private void swap(int[] nums,int a,int b){
int temp=nums[a];
nums[a]=nums[b];
nums[b]=temp;
}
}
public class Solution {
public void nextPermutation(int[] nums) {
int len = nums.length;
boolean flag = false;
for(int i=len-1; i>=1; i--){
if(nums[i]>nums[i-1]){
int j=i;
while(j<len-1){
if(nums[j+1]<=nums[i-1] ){
break;
}
j++;
}
if(nums[j]>nums[i-1]){
swap(nums,i-1,j);
}
reverse(nums,i,len-1);
flag = true;
break;
}
}
if(!flag){
reverse(nums,0,len-1);
}
}
private void reverse(int[] nums,int i,int j){
while(i<j){
swap(nums,i++,j--);
}
}
private void swap(int[] nums,int a,int b){
int temp=nums[a];
nums[a]=nums[b];
nums[b]=temp;
}
}
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