买卖股票的最佳时期(121)
- class Solution {
- public int maxProfit(int[] prices) {
- int res = 0;
- int min = Integer.MAX_VALUE;
- for (int i = 0; i < prices.length; i++){
- min = Math.min(min, prices[i]);
- res = Math.max(res,prices[i] - min);
- }
- return res;
- }
- }
复制代码 贪心就是贪局部最优解, 扩散到全局
跳跃游戏(055)
- class Solution {
- public boolean canJump(int[] nums) {
- int max_length = 0;
- int i = 0;
- for (; max_length >= i && i < nums.length; i++){
- max_length = Math.max(max_length, i + nums[i]);
- }
- return i == nums.length;
- }
- }
复制代码 max_length来维护理论可达距离
跳跃游戏II(045)
- class Solution {
- public int jump(int[] nums) {
- int n = nums.length;
- int res = 0;
- int curr = 0;
- int nextCurr = 0;
- for (int i = 0; i < n-1 ; i++){
- nextCurr = Math.max(nextCurr, i+ nums[i]);
- if (i == curr){
- curr = nextCurr;
- res++;
- }
- }
- return res;
- }
- }
复制代码 curr 维护本次跳跃最大可达距离
nextCurr 通过遍历途经点, 维护下次跳跃最大可达距离
划分字母区间(763)
- class Solution {
- public List<Integer> partitionLabels(String s) {
- int n = s.length();
- char[] charArray = s.toCharArray();
- int[] last = new int[26];
- for (int i = 0 ; i < n; i++){
- last[charArray[i]- 'a'] = i;
- }
- List<Integer> res = new ArrayList<>();
- int start = 0;
- int end = 0;
- for (int i = 0; i < n; i++){
- end = Math.max(end, last[charArray[i] - 'a']);
- if (end == i){
- res.add(end - start + 1);
- start = i+1;
- }
- }
- return res;
- }
- }
复制代码 将字符串预处理, 产生每个字符的最大索引
提取[start,end]范围内字符的最远索引来更新end
遇到这种熟悉又陌生的题型真别怕, 先把陌生数据转换成熟悉的, 这题就跟跳跃游戏II(045)一样了
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