303. Range Sum Query - Immutable
用一个数组保存从0到当前位置的和。
class NumArray {
int[] sumnums;
public NumArray(int[] nums) {
if(nums.length==0)
return;
sumnums = new int[nums.length];
sumnums[0] = nums[0];
for(int i=1;i<nums.length;i++){
sumnums[i] = sumnums[i-1] + nums[I];
}
}
public int sumRange(int i, int j) {
if(i==0)
return sumnums[j];
return sumnums[j] - sumnums[i-1];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
304. Range Sum Query 2D - Immutable
借鉴303题的思路,同样构造一个求和的矩阵,行列关系共分为四种,需要仔细分析每种情况下的计算:
class NumMatrix {
int[][] summatrix;
public NumMatrix(int[][] matrix) {
if(matrix==null || matrix.length==0)
return;
summatrix = new int[matrix.length][matrix[0].length];
for(int i=0;i<matrix.length;i++){
for(int j=0;j<matrix[i].length;j++){
if(i==0 && j==0)
summatrix[i][j] = matrix[i][j];
else if(i==0)
summatrix[i][j] = summatrix[i][j-1] + matrix[i][j];
else if(j==0)
summatrix[i][j] = summatrix[i-1][j] + matrix[i][j];
else
summatrix[i][j] = summatrix[i-1][j] + summatrix[i][j-1] - summatrix[i-1][j-1] + matrix[i][j];
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
if(row1==0 && col1==0)
return summatrix[row2][col2];
else if(row1==0)
return summatrix[row2][col2] - summatrix[row2][col1-1];
else if(col1==0)
return summatrix[row2][col2] - summatrix[row1-1][col2];
else
return summatrix[row2][col2] - summatrix[row1-1][col2] - summatrix[row2][col1-1] + summatrix[row1-1][col1-1];
}
}
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix obj = new NumMatrix(matrix);
* int param_1 = obj.sumRegion(row1,col1,row2,col2);
*/
307. Range Sum Query - Mutable
这里同303题,用另一个数组保存了原数组中每个元素的值,每次更新的时候,不要忘记将该位置对应元素的值进行更新。
class NumArray {
int[] sumnums;
int[] orgnums;
public NumArray(int[] nums) {
if(nums==null || nums.length==0)
return;
sumnums = new int[nums.length];
sumnums[0] = nums[0];
orgnums = new int[nums.length];
orgnums[0] = nums[0];
for(int i=1;i<nums.length;i++){
sumnums[i] = sumnums[i-1] + nums[I];
orgnums[i] = nums[I];
}
}
public void update(int i, int val) {
for(int j=i;j<sumnums.length;j++){
sumnums[j] += (val - orgnums[I]);
}
orgnums[i] = val;
}
public int sumRange(int i, int j) {
if(i==0)
return sumnums[j];
return sumnums[j] - sumnums[i-1];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/
309. Best Time to Buy and Sell Stock with Cooldown
使用回溯法,超时。。
class Solution {
int maxProfit = 0;
public int maxProfit(int[] prices) {
backprop(prices,0,0,0,0);
return maxProfit;
}
public void backprop(int[] prices,int start,int profit,int buyPrice,int type){
if(start >= prices.length){
maxProfit = Math.max(maxProfit,profit);
}
if(type==0){
for(int i=start;i<prices.length;i++){
backprop(prices,i+1,profit,prices[i],1);
}
}
else if(type==1){
for(int i=start;i<prices.length;i++){
backprop(prices,i+2,profit + (prices[i]-buyPrice),0,0);
}
}
}
}
此题需要维护三个一维数组buy, sell,和rest。其中:
buy[i]表示在第i天之前最后一个操作是买,此时的最大收益。
sell[i]表示在第i天之前最后一个操作是卖,此时的最大收益。
rest[i]表示在第i天之前最后一个操作是冷冻期,此时的最大收益。
我们写出递推式为:
buy[i] = max(rest[i-1] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
rest[i] = max(sell[i-1], rest[i-1])
解读一下,第一个公式buy[i],如果第i天可以买,那么i-1天会是冷冻期,否则,不能进行购买,因此比较rest[i-1] - price和 buy[i-1]的大小。第二个公式sell[i],可以卖的条件是,i-1天之前最后一个操作时买入股票,否则,不能卖出,因此比较buy[i-1] + price和sell[i-1]。第三个式子,冷冻期的可能性有两个,一个是昨天刚进行卖的交易,另一个是继续冷冻期。
上述递推式很好的表示了在买之前有冷冻期,买之前要卖掉之前的股票。一个小技巧是如何保证[buy, rest, buy]的情况不会出现,这是由于buy[i] <= rest[i], 即rest[i] = max(sell[i-1], rest[i-1]),这保证了[buy, rest, buy]不会出现。
另外,由于冷冻期的存在,我们可以得出rest[i] = sell[i-1],这样,我们可以将上面三个递推式精简到两个:
buy[i] = max(sell[i-2] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
因此,代码如下:
class Solution {
public int maxProfit(int[] prices) {
if(prices==null || prices.length <2)
return 0;
int[] buy = new int[prices.length];
int[] sell = new int[prices.length];
buy[0] = -prices[0];
buy[1] = Math.max(-prices[0],-prices[1]);
sell[0] = 0;
sell[1] = Math.max(0,prices[1]-prices[0]);
for(int i=2;i<prices.length;i++){
buy[i] = Math.max(sell[i-2] - prices[i],buy[i-1]);
sell[i] = Math.max(buy[i-1] + prices[i],sell[i-1]);
}
return sell[prices.length-1];
}
}
310. Minimum Height Trees
答案只可能有一个或两个节点。思路为依次删除叶子节点,剩下的1/2个节点即为解。
速度问题:如果每次遍历选出叶子节点,速度比较慢,可以每次删除当前叶子节点时,将新的叶子节点记录下来。
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> leaves = new ArrayList<>();
if(n <= 1) {
return Collections.singletonList(0);
}
Map<Integer, Set<Integer>> graph = new HashMap<>(); // list of edges to Ajacency Lists
for(int i = 0; i < n; i++) {
graph.put(i, new HashSet<Integer>());
}
for(int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
}
for(int i = 0; i < n; i++) {
if(graph.get(i).size() == 1) {
leaves.add(i);
}
}
while(n > 2) {
n -= leaves.size();
List<Integer> newLeaves = new ArrayList<>();
for(int leaf : leaves) {
for(int newLeaf : graph.get(leaf)) {
graph.get(leaf).remove(newLeaf);
graph.get(newLeaf).remove(leaf);
if(graph.get(newLeaf).size() == 1) {
newLeaves.add(newLeaf);
}
}
}
leaves = newLeaves;
}
return leaves;
}
}
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