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1. 把整个图色看成是一棵树，然后dfs
根节点有三个分叉：r,g,b
第二层(g,b),(r,b),(r,g)
所以就是相当于path sum
1 int min = Integer.MAX_VALUE; 2 3 public int minCost(int[][] costs) { 4 if(costs.length == 0 || costs[0].length == 0) { 5 return 0; 6 } 7 for(int i = 0; i < 3; i++) { 8 sum(costs, 0, 0, i); 9 } 10 return min; 11 } 12 13 private void sum(int[][] costs, int houseNo, int lastSum, int curColor) { 14 if(houseNo == costs.length - 1) { 15 min = Math.min(min, lastSum + costs[houseNo][curColor]); 16 return; 17 } 18 List<Integer> list = new ArrayList<Integer>(Arrays.asList(0,1,2)); 19 list.remove(curColor); 20 for(int i: list) { 21 sum(costs, houseNo + 1, lastSum + costs[houseNo][curColor], i); 22 } 23 }
时间复杂度应该是O(2^n)
哈哈然后LTE了
2. 动规
既然递归会超时，那就动规了
1 public int minCost(int[][] costs) { 2 int len = costs.length; 3 if(len == 0 || costs.length == 0) { 4 return 0; 5 } 6 int[][] path = new int[len][3]; 7 int min = Integer.MIN_VALUE; 8 for(int i = 0; i < 3; i++) { 9 path[0][i] = costs[0][i]; 10 } 11 for(int i = 1; i < len; i++) { 12 path[i][0] = Math.min(path[i-1][1], path[i-1][2]) + costs[i][0]; 13 path[i][1] = Math.min(path[i-1][0], path[i-1][2]) + costs[i][1]; 14 path[i][2] = Math.min(path[i-1][0], path[i-1][1]) + costs[i][2]; 15 } 16 return Math.min(path[len-1][0], Math.min(path[len-1][1], path[len-1][2])); 17 }
这样只有O(N)了
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原文地址：http://www.cnblogs.com/warmland/p/5759538.html
