Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1....
If all digits are sorted in ascending order, then we need to swap last two digits. For example, 1234. For other cases, we need to process the number from rightmost side (why? because we need to find the smallest of all greater numbers) Now the main algorithm works in following steps -...
The next greater element for 4 is 5 and the next greater element for elements 3, 2, 1 is -1 as there is no greater element after them.AlgorithmInitialise the array with random numbers. Initialise a stack. Add first element to the stack. Iterate through the element of the array. If ...
Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater n...
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Otherwise, the function returns false to indicate that the arrangement is not greater than the previous, but the lowest possible (sorted in ascending order).Example 123456789101112131415161718 // next_permutation example #include <iostream> // std::cout #include <algorithm> // std::next_permutation...
The range used is [first, last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. return value: true : if the function could rearrange the object as a lexicographicaly greater permutation. Otherwise, the ...
( int elem1, int elem2 ) { if ( elem1 < 0 ) elem1 = - elem1; if ( elem2 < 0 ) elem2 = - elem2; return elem1 < elem2; }; int main( ) { // Reordering the elements of type CInt in a deque // using the prev_permutation algorithm CInt c1 = 5, c2 = 1, c3 = ...
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>&nums) { stack<int>s; map<int,int>mp;for(inti =0;i < nums.size(); i++){while(!s.empty() && s.top() <nums[i]){ mp[s.top()]=nums[i]; s.pop();
publicintnextGreaterElement(intn) { char[] number = (n +"").toCharArray(); inti, j; // I) Start from the right most digit and // find the first digit that is // smaller than the digit next to it. for(i = number.length-1; i >0; i--) ...