給自己的目標：[LeetCode](https://leetcode.com/ "Online Judge Platform") 上每日一題
在做題的過程中記錄下解題的思路或者重要的代碼碎片以便后來翻閱。
項目源碼：github上的Leetcode
31. Next Permutation
題目：給出一個序列，尋找比當前排列順序大的下一個排列。規則如下：
1,2,3 ->next
1,3,2 ->next
2,1,3 ->next
2,3,1 ->next
3,1,2 ->next
3,2,1 ->next
1,2,3 ->loop
因為是找遞增的下一個排列，所以從后往前找到第一個升序對的位置，如1,2,4,3,1， 從后向前找就是2,4,3,1，因為2比前一個數4小，所以就鎖定2這個數。
之后就是在4,3,1中找到比2大的最小的那個數3，將3與2對換得到降序排列4,2,1.然后就是將4,2,1反序得到1,2,4.最終結果就是1,3,1,2,4
public class Solution {
public void nextPermutation(int[] nums) {
if (nums.length == 0 || nums.length == 1) return;
int len = nums.length;
int index = len - 1;
int value = nums[index];
for (index = index - 1; index >= 0; index--) {
if (nums[index] < value) {
value = nums[index];
break;
}
value = nums[index];
}
if (index < 0) {
reversal(nums, 0, len - 1);
} else {
for (int j = len - 1; j > index; j--) {
if (nums[j] > value) {
nums[index] = nums[j];
nums[j] = value;
reversal(nums, index + 1, len - 1);
break;
}
}
}
}
public void reversal(int[] nums, int start, int end) {
int len = end + 1 - start;
for (int i = 0; i < len / 2; i++) {
int k = nums[start + i];
nums[start + i] = nums[end - i];
nums[end - i] = k;
}
}
}
32. Longest Valid Parentheses
題目：輸入一串只包含"(",")"的字符串，求合法子串的最大值。
")()())", where the longest valid parentheses substring is "()()", which has length = 4.
一道很容易TLE的題目，最優解法是使用 stack 記錄字符串中"("出現時的index，當出現")"進行匹配時相減得到長度。
public class Solution {
public int longestValidParentheses(String s) {
if (s.isEmpty()) return 0;
Stack<Integer> ms = new Stack<>();
int maxlen = 0;
int last = -1; // Position of the last unmatched ')'
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') ms.push(i); //
else if (ms.empty()) last = i; // == ')' and the stack is empty, it means this is a non-matching ')'
else {
ms.pop(); // pops the matching '('.
if (ms.empty()) // If there is nothing left in the stack, it means this ')' matches a '(' after the last unmatched ')'
maxlen = Math.max(maxlen, i - last);
else // otherwise,
maxlen = Math.max(maxlen, i - ms.peek());
}
}
return maxlen;
}
}
33.Search in Rotated Sorted Array
題目：給出一組有序的數組，如 0 1 2 4 5 6 7，將其分成兩段并互換，比如會成為4 5 6 7 0 1 2。給出一個 target 值，求在數組中的位置，若不存在輸出-1
一種最簡單的方法是直接遍歷過，時間復雜度為O(n),另一種是先使用 binary search 找出旋轉點（最小值）所在的位置，之后在用二分遍歷一次獲得 target 所在的位置。
/**
* 簡單遍歷
*/
public class Solution {
public int search(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
if (nums[i] == target) {
return i;
}
}
return -1;
}
}
34. Search for a Range
題目：給出一個有序的數組和一個目標值，求該目標值在數組的范圍（由于目標值在數組中存在好幾個）。要求時間復雜度為O(logn)
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
一個很簡單的二分查找題，要注意的是即使找到一個目標值也要繼續向左右查找下去以便確定目標值在數組中的范圍。
public class Solution {
public int[] searchRange(int[] nums, int target) {
int start = -1, end = -1;
int st = 0, len = nums.length - 1;
while (st <= len) {
int mid = (len + st + 1) / 2;
if (nums[mid] == target) {
start = binarySearch(nums, st, mid - 1, target, true);
end = binarySearch(nums, mid + 1, len, target, false);
if (start == -1) start = mid;
if (end == -1) end = mid;
break;
} else if (nums[mid] > target) {
len = mid - 1;
} else if (nums[mid] < target) {
st = mid + 1;
}
}
return new int[]{start, end};
}
public int binarySearch(int[] nums, int start, int end, int target, boolean left) {
if (start > end) {
return -1;
}
int mid = (start + end + 1) / 2;
int index = -1;
if (nums[mid] == target) {
if (left) {
index = binarySearch(nums, start, mid - 1, target, left);
} else {
index = binarySearch(nums, mid + 1, end, target, left);
}
if (index == -1) {
return mid;
}
return index;
} else if (nums[mid] > target) {
return binarySearch(nums, start, mid - 1, target, left);
} else if (nums[mid] < target) {
return binarySearch(nums, mid + 1, end, target, left);
} else {
return index;
}
}
}
35. Search Insert Position
題目：給出一組有序數組和一個目標數字，求目標數字在數組中的位置，沒找到時則求插入數組的位置。
Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
一道簡單的二分查找題
public class Solution {
public int searchInsert(int[] nums, int target) {
int len = nums.length;
int start = 0;
int end = len - 1;
int index = start;
while (start <= end) {
int mid = (start + end) / 2;
if (target > nums[mid]) {
start = mid + 1;
index = start;
} else if (target < nums[mid]) {
end = mid - 1;
index = mid;
} else {
return mid;
}
}
return index;
}
}
