There is a fence with n posts, each post can be painted with one of the k colors.
You have to paint all the posts such that no more than two adjacent fence posts have the same color.
Return the total number of ways you can paint the fence.
Note:?n and k are non-negative integers.

涂柵欄條最多兩個(gè)顏色在一起。艾瑪， 鬧心， 二刷看答案竟然還要想一會(huì)兒！！！尼瑪啊?

當(dāng)然這道題得用動(dòng)態(tài)規(guī)劃， 將1，2柵欄條相同顏色和不同顏色區(qū)分開， 分別計(jì)算可能性， 返回總和。還要考慮異常條件 沒有柵欄和僅有一個(gè)柵欄的情況。

空間復(fù)雜度O(2n)， 根據(jù)規(guī)則可以發(fā)現(xiàn)只要另個(gè)變量O(2) 就可以做到。修改代碼。

public int numWays(int n, int k) {
? ? ? if(n == 0){
? ? ? ? ? ?return 0;
? ? ? }
? ? ? if(n == 1){
? ? ? ? ? ? return k;
? ? ? }
? ? ? int[] dp_same = new int[n];
? ? ? int[] dp_diff = new int[n];
? ? ? dp_same[0] = dp_same[1] = k;
? ? ? dp_diff[0] = k;
? ? ? dp_diff[1] = k*(k-1);
? ? ? for(int i = 2; i < n; i++ ){
? ? ? ? ? ? dp_same[i] = dp_diff[i-1];
? ? ? ? ? ? dp_diff[i] = (k - 1)*(dp_diff[i-1] + dp_same[i-1]);
? ? ? }
? ? ? return dp_diff[n-1] + dp_same[n-1];
}

public int numWays(int n, int k) {
? ? ? if(n == 0){
? ? ? ? ? ?return 0;
? ? ? }
? ? ?if(n == 1){
? ? ? ? ? ? return k;
? ? ?}
? ? ?int same = k;
? ? ?int diff = k * (k-1);
? ? ?for(int i = 2; i < n; i++){
? ? ? ? ? int temp = diff;
? ? ? ? ? diff = (same+diff)*(k-1);
? ? ? ? ? same = temp;
? ? ?}
? ? ?return same + diff;
}