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problems/0198.打家劫舍.md
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@ -141,7 +141,36 @@ class Solution {
}
}
// 空间优化 dp数组只存与计算相关的两次数据
// 使用滚动数组思想，优化空间
// 分析本题可以发现，所求结果仅依赖于前两种状态，此时可以使用滚动数组思想将空间复杂度降低为3个空间
class Solution {
public int rob(int[] nums) {
int len = nums.length;
if (len == 0) return 0;
else if (len == 1) return nums[0];
else if (len == 2) return Math.max(nums[0],nums[1]);
int[] result = new int[3]; //存放选择的结果
result[0] = nums[0];
result[1] = Math.max(nums[0],nums[1]);
for(int i=2;i<len;i++){
result[2] = Math.max(result[0]+nums[i],result[1]);
result[0] = result[1];
result[1] = result[2];
}
return result[2];
}
}
// 进一步对滚动数组的空间优化 dp数组只存与计算相关的两次数据
class Solution {
public int rob(int[] nums) {
if (nums.length == 1) {
@ -151,11 +180,11 @@ class Solution {
// 优化空间 dp数组只用2格空间 只记录与当前计算相关的前两个结果
int[] dp = new int[2];
dp[0] = nums[0];
dp[1] = nums[0] > nums[1] ? nums[0] : nums[1];
dp[1] = Math.max(nums[0],nums[1]);
int res = 0;
// 遍历
for (int i = 2; i < nums.length; i++) {
res = (dp[0] + nums[i]) > dp[1] ? (dp[0] + nums[i]) : dp[1];
res = Math.max((dp[0] + nums[i]) , dp[1] );
dp[0] = dp[1];
dp[1] = res;
}