rob[i] 代表在 i 处最大能偷到的钱。Time = O(n), space = O(n).
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
return Math.max(nums[0], nums[1]);
}
int[] rob = new int[nums.length];
rob[0] = nums[0];
rob[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < nums.length; i++) {
rob[i] = Math.max(rob[i - 2] + nums[i], rob[i - 1]);
}
return rob[rob.length - 1];
}
}
Optimize space to O(1)：
class Solution {
public int rob(int[] nums) {
int rob0 = 0;
int rob1 = 0;
int rob = 0;
for (int i = 0; i < nums.length; i++) {
rob = Math.max(rob0 + nums[i], rob1);
rob0 = rob1;
rob1 = rob;
}
return rob;
}
}