intN = nums.length; int[] dp =newint[N+1]; dp[0] =0; intres = Integer.MIN_VALUE; for(intk =1; k <= N; k++) { dp[k] = Math.max(dp[k-1],0) + nums[k-1]; res = Math.max(res, dp[k]); } returnres; } 时间复杂度,空间复杂度。 动态规划的空间优化 上面的动态规划...
(nums.length==2) return Math.max(nums[begin],nums[begin+1]); int[] dp=new int[end]; dp[begin]=nums[begin];//这里不要用0、1,这里用begin、end dp[begin+1]=Math.max(nums[begin],nums[begin+1]); for(int i=2;i<end;i++){ dp[i]=Math.max(dp[i-1],dp[i-2]+nums[i]);...
public static int getNum(int[] nums) { if (nums == null || nums.length == 0) return 0; int target = 0; int n = nums.length; for (int i = 0; i < nums.length; ++i) target += nums[i]; boolean[][] dp = new boolean[n + 1][target + 1]; dp[0][0] = true; int...
public int maxSumAfterOperation(int[] nums) { int n = nums.length; int[][] dp = new int[n][2]; dp[0][0] = nums[0]; dp[0][1] = nums[0] * nums[0]; int ans = dp[0][1]; for (int i = 1; i < n; i++) { ...
intn=weightslength;jixiangip.com; int[][]dp=newint[n+1][capacity+1]; for(inti=1;i<=n;i++){ for(intj=1;j<=capacity;j++){ if(weights[i-1]>j){ dp[i][j]=dp[i-1][j]; }else{ dp[i][j]=Mathmax(dp[i-1][j],dp[i-1][j-weights[i-1]]+values[i-1]); ...
}switchvx := (*x).(type) {case*big.Int:varz *big.Int= big.NewInt(0)returnwrap(z.Lsh(vx, amount)) }panic("bad type") } 开发者ID:weinholt,项目名称:conscheme,代码行数:28,代码来源:arithmetic.go 示例4: _string_to_number
求城市x到城市y的最短距离。 答案2022-03-11: 有向图,没有负数环。小根堆。 代码用golang编写。代码如下: ```go package main import ( "fmt" "math" "sort" ) func main() { roads := [][]int{{1, 2, 3}, {1, 3, 5}, {2, 3, 1}} ...
public int wiggleMaxLength(int[] nums) { int n = nums.length; int[][] dp = new int[n][2]; int ans = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < i; j++) { if (nums[i] > nums[j]) { dp[i][0] = Math.max(dp[i][0],dp[j][1] + 1); ...
[] dp=new int[nums.length()+1]; Arrays.fill(dp,1);//初始化为1,有1种译码方式 for(int i=2;i<nums.length()+1;i++){ //11-19 21-26 if(((nums.charAt(i-2)=='1'&&nums.charAt(i-1)!='0')||(nums.charAt(i-2)=='2'&&nums.charAt(i-1)<'7'&& nums.charAt(i-1)>'0'...
if(nums[i] < min) min = nums[i]; } returnnewint[]{max, min}; } // 返回最大值和第二大值 int[]max1_and_max2(int[] nums) { intn = nums.length; intmax1, max2; max1 = max2 = Integer.MIN_VALUE; for(inti =0; i < n; i++) { ...